Infinite sequences and series help

In summary: The sequence that makes up the terms in the series.2. The sequence of partial sums.Using the first example above, the sequence of terms would be{2/8, 4/11, 6/14, ..., 2n/(3n + 5), ...}The seqence of partial sums would be{2/8, 2/8 + 4/11, 2/8 + 4/11 + 6/14, ...}(I haven't bothered to simplify the fractions or combine them.)A series converges if its sequence of partial sums converges.In summary, the conversation discusses the concept of limit and summation in relation to sequences
  • #1
christian0710
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Hi I don't understand the logic in the picture i added. They say that "that sum of the series = the limit of the sequence"

The limit is 2/3 BUT the sum, Ʃ, must be 2*1/(3*1+5) + (2*2/(2*3+5) + 2*3/(2*3+5) ...+
Which is obviously much larger than 2/3 if all the terms are added together?? it's like adding 2/3+2/3+2/3 which is = 6/2 =2??

Is there something I'm misunderstanding here?
 

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  • #2
what is the a sub n term? the a sub n terms are added together to get the s sub n

the sequence referred to is the s sub n terms as defined by the 2*n/(3*n+5) you don't sum these. Instead you must evaluate the term as n approaches infinity.

so that when n is very large the sequence terms become 2*n / 3*n and the n drops out to leave 2/3
 
  • #3
Well if we have the series Ʃan= 1/2n i can see that the sum would add up to one.

Ahhh So the limit of one Lim 1 = 1 = the sum?? Is that how I should understand it?
 

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  • #4
Ahh so the partial sequence of Sn adds up to 2n/(3n+5) as you said (just like 1/2^2 adds up to one). so if we add up all the terms of some imaginary sequence we would get the sum to be 2n/(3n+5), so the sum is the end result and NOT 2*1/(3*1+5) + (2*2/(2*3+5) + 2*3/(2*3+5) ...+ which is the result of adding up the sums.

So the limit is the the sum because the sum is 2n/(3n+5) and when n gets big it's 2/3. Is that correctly understood? And thank you :)
 
  • #5
So is it correctly understood that the limit of a sequence 1/2^n is zero but the limit of the partial sums of that sequence is 1? That's the diffrence?
 
  • #6
christian0710 said:
Ahh so the partial sequence of Sn [STRIKE]adds [/STRIKE]up to 2n/(3n+5) as you said (just like 1/2^2 adds up to one). so if we add up all the terms of some imaginary sequence we would get the sum to be 2n/(3n+5), so the sum is the end result and NOT 2*1/(3*1+5) + (2*2/(2*3+5) + 2*3/(2*3+5) ...+ which is the result of adding up the sums.

So the limit is the the sum because the sum is 2n/(3n+5) and when n gets big it's 2/3. Is that correctly understood? And thank you :)

Don't say adds up, its not adding up its a sequence that approaches a limit, the limit is 2/3

Look at a simpler example:

Ʃ(2n+1) = n^2

an = (2n+1)

sn = n^2

so the sequence of sn is { 1, 4, 9, 16, 25 ... }

so as n approaches infinity what is the last term of the sequence of sn
 
  • #7
don't look at it that way

divide each term by the highest power variable and take each portion of the equations limit, separately.

a quick way of doing it is to take a ratio of coefficients of variables in this case of an
 
  • #8
Hmm well Okay first, Ʃ(2n+1) means 2*1+1 +2*2+1 ...(2n+1) this is the action of adding up a sequence of numbers, right? Isn't that why we use a summation symbol? Isn't it the definition of summation "Sum" = add up?

Now follow me here (and please tell me it's correctly understood :-) then I'll be happy )
The sum of this series 1/2+1/4+1/8...+ has a pattern 1/2^n and we denote it Ʃ(1/2^n ), By taking a partial sum Sn= 1/2+1/4+1/8 We check if this sum approaches a limit (in this case Lim Sn=1)
So the sum of the series Ʃ(1/2^n ) is the limit of the sequence of partial sums, right??

The limit of 1/2^n approaches zero but the limit of the partial sum Sn approaches one.

This must be correct, please :D
 
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  • #9
With respect to your prior statement, it would be better to say:

partial sum of Sn is given by 2n/(3n+5) and this approaches 2/3 as n approaches infinity.

perhaps we're splitting hairs here but I didn't want you to think that the 2n/(3n+5)

is the same as s1+s2+s3...
christian0710 said:
The sum of this series 1/2+1/4+1/8...+ has a pattern 1/2^n and we denote it Ʃ(1/2^n ),

By taking a partial sum S1=1/2, S2=3/4, S3= 1/2+1/4+1/8 aand Sn= Ʃ(1/2^n ),

and we get the sequence of partial sums:

{1/2, 3/4, 7/8...}
We check if this sum approaches a limit (in this case Lim Sn=1)

So the sum of the series Ʃ(1/2^n ) is the limit of the sequence of partial sums, right??

The limit of 1/2^n approaches zero but the limit of the partial sum Sn approaches one.

This must be correct, please :D

With respect to this statement it appears to be correct.
 
  • #10
Thank you! :)

I see because you wanted me to understand that 2n/(3n+5) is the infinite series (when n goes to infinity) and s1+s2+s3... is a sequence of numbers.
 
  • #11
christian0710 said:
Thank you! :)

I see because you wanted me to understand that 2n/(3n+5) is the infinite series (when n goes to infinity)
The infinite series is
$$ \sum_{n = 1}^{\infty} \frac{2n}{3n + 5}$$

2n/(3n + 5) is merely the n-th term in the series or the n-th term in the sequence that underlies the series.
christian0710 said:
and s1+s2+s3... is a sequence of numbers.
No, s1 + s2 + s3 +... is a sum (or infinite series). The ellipsis (...) indicates that this sum continues indefinitely, following the same pattern.

Every infinite series involves two sequences:
1. The sequence that makes up the terms in the series.
2. The sequence of partial sums.

Using the first example above, the sequence of terms would be
{2/8, 4/11, 6/14, ..., 2n/(3n + 5), ...}
The seqence of partial sums would be
{2/8, 2/8 + 4/11, 2/8 + 4/11 + 6/14, ...}

(I haven't bothered to simplify the fractions or combine them.)

A series converges if its sequence of partial sums converges.
 
  • #12
Mark44 said:
The infinite series is
$$ \sum_{n = 1}^{\infty} \frac{2n}{3n + 5}$$

2n/(3n + 5) is merely the n-th term in the series or the n-th term in the sequence that underlies the series.

No, s1 + s2 + s3 +... is a sum (or infinite series). The ellipsis (...) indicates that this sum continues indefinitely, following the same pattern.

Every infinite series involves two sequences:
1. The sequence that makes up the terms in the series.
2. The sequence of partial sums.

Using the first example above, the sequence of terms would be
{2/8, 4/11, 6/14, ..., 2n/(3n + 5), ...}
The seqence of partial sums would be
{2/8, 2/8 + 4/11, 2/8 + 4/11 + 6/14, ...}

(I haven't bothered to simplify the fractions or combine them.)

A series converges if its sequence of partial sums converges.

Thank you I finally understand Sn now completely,
S1 = a1
S2= a1 +a1
S3 = a1+a2+a3.

So the n-th term of an is 2n/(3n + 5) and the n-th term of Sn as i goes from 1 to n gives approaches the limit/sum for the finite series if it converges.
 
  • #13
But wait, this part: 2n/(3n + 5) is merely the n-th term in the series or the n-th term in the sequence that underlies the series.

What's the difference between the series and the sequence that underlies the series?
The series is the infinite sequence of numbers right? and the sequence is a part of that series?
 
  • #14
And to this "1. The sequence that makes up the terms in the series."
A term is 2/8 , so don't you mean "the terms that make up the sequence (a1 +a2+a3) in the series"
 
  • #15
christian0710 said:
But wait, this part: 2n/(3n + 5) is merely the n-th term in the series or the n-th term in the sequence that underlies the series.

What's the difference between the series and the sequence that underlies the series?
The series is the infinite sequence of numbers right? and the sequence is a part of that series?

The series is what you get when you add all of the terms in the sequence.
The sequence is a list, albeit an infinitely long list: {a1, a2, a3 ,,, an...}
The series (AKA infinite series) is the sum that you get when you add all of the terms in the sequence.

christian0710 said:
And to this "1. The sequence that makes up the terms in the series."
A term is 2/8 , so don't you mean "the terms that make up the sequence (a1 +a2+a3) in the series"

I think you are confusing the sequence that represents the terms, which is abbreviated as {an}, with the sequence of partial sums. {S1, S2, S3, ... , Sn, ...}

Here S1 = a1
S2 = a1 + a2 = S1 + a2
S3 = a1 + a2 + a3= S2 + a3
and so on.

In general, Sn = ## \sum_{i = 1}^n a_n##, which is a finite sum.
 
  • #16
I'm sory, You are right I'm confusing things together. I sat down, re-read everything, and here it is. I understand it now, I really spent time thinking about it.

We have a sequence of number fx the sequence 1/2n, and we can test the limit of such a sequence, by doing this we want to find out what the sequence approaches as n → ∞
Lim1/2n = 0
If we add the terms of an infinite sequence we get what we call an Infinite series. The series is a number/sum denoted Ʃ(an) consisting of the addition of the numbers in the sequence an.

Instead of adding the whole sequence to infinity, we add a part of the sequence. We call this The partial Sums, denoted Sn.
Sn = A1+a2+a3…an(…=infinity)
We can test if this series of numbers converges toward a finite number, or if it diverges to infinity by using partial sum.

Example
If we have a sequence an = 1/2n
The sequence is: 1/2 ,1/4, 1/8…,1/nn...(...= going to infinity)

The limit of the sequence if we let n go toward infinity is Lim 1/2n =0

On the other hand The series Ʃ(1/2n) also has a limit. We can prove this by a rule that says if the Sequece of the partial sum {Sn} is convergent (which it is, because the limit of the sequence {Sn}=1/2 ,1/2+1/4,1/2+1/4+ 1/8…,1/nn is 1} then the series Ʃ(1/2n) is convergent, and it is because it converges toward 1.
 
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  • #17
You were right, I was confusing the sequence of terms with the sequence of partial sums, that was what confused me so much in my calc book. Thank you! :)
 
  • #18
christian0710 said:
I'm sory, You are right I'm confusing things together.
No need for apology. This is complicated stuff when you're first exposed to it.
christian0710 said:
I sat down, re-read everything, and here it is. I understand it now, I really spent time thinking about it.
I can tell. It looks like you understand pretty well.

There are a couple of nits below, but otherwise you seem to have a good handle on the ideas.
christian0710 said:
We have a sequence of number fx the sequence 1/2n, and we can test the limit of such a sequence, by doing this we want to find out what the sequence approaches as n → ∞
Lim1/2n = 0
If we add the terms of an infinite sequence we get what we call an Infinite series. The series is a number/sum denoted Ʃ(an) consisting of the addition of the numbers in the sequence an.
If the series is convergent, then it represents a number. Other series are divergent, so they don't represent a number.
christian0710 said:
Instead of adding the whole sequence to infinity, we add a part of the sequence. We call this The partial Sums, denoted Sn.
Sn = A1+a2+a3…an(…=infinity)
Sn represents the sum of a finite number of terms for each value of n.
Sn = a1 + a2 + ... + an. However, as n increases, the number of terms that make up Sn gets larger, but there are always n terms.
christian0710 said:
We can test if this series of numbers converges toward a finite number, or if it diverges to infinity by using partial sum.
The sequence of partial sums doesn't have to grow unboundedly large for the series to diverge. For example, in this series -- ##\sum_{i = 1}^{\infty} (-1)^n## -- the sequence of partial sums is {-1, 0, -1, 0, -1, 0, ...}. Notice that I am actually adding the terms in the series to get this sequence of partial sums: {-1, -1 + 1, -1 + 1 - 1, ...}. This sequence (of partial sums) is not growing larger, but the terms in the sequence never settle into a particular value. Hence ##\sum_{i = 1}^{\infty} (-1)^n## is divergent.
christian0710 said:
Example
If we have a sequence an = 1/2n
The sequence is: 1/2 ,1/4, 1/8…,1/nn...(...= going to infinity)
The last term that you show should be 1/2n.
christian0710 said:
The limit of the sequence if we let n go toward infinity is Lim 1/2n =0

On the other hand The series Ʃ(1/2n) also has a limit. We can prove this by a rule that says if the Sequece of the partial sum {Sn} is convergent (which it is, because the limit of the sequence {Sn}=1/2 ,1/2+1/4,1/2+1/4+ 1/8…,1/nn is 1}
The last term that you show above should be 1/2 + 1/4 + 1/8 + ... + 1/2n.

A little work shows that the sequence of partial sums can be written as {1/2, 3/4, 7/8, ... , (2n - 1)/22, ...}
christian0710 said:
then the series Ʃ(1/2n) is convergent, and it is because it converges toward 1.
We say "converges to 1", not "toward".

Good work!
 
  • #20
Man, I really appreciate all your help. You are great at explaining this topic!

So If if the sequence of a partial sum is (2n - 1)/22, Then this must mean that there is a pattern, just like for a sequence an and for a series Ʃan. So if the partial sum Sn = (2n - 1)/22 what exactly does that tell us?

It must be the partial sum of the n-th term in the sequence an,

I remember from the text "The sum of a series is the limit of the sequence of partial sums", so the sum of the series Ʃ1/2n must be the limit of Sn? But does it make sense to speak about a limit of Sn = (2n - 1)/22 which is a finite number? It's like saying "the limit of 25 = 25" right?
 
  • #22
christian0710 said:
Man, I really appreciate all your help. You are great at explaining this topic!
Corrected a couple of your subscripts that should be exponents in the following.
christian0710 said:
So If if the sequence of a partial sum is (2n - 1)/22, Then this must mean that there is a pattern, just like for a sequence an and for a series Ʃan. So if the partial sum Sn = (2n - 1)/22 what exactly does that tell us?
It gives you a handy, dandy form that is ripe for you to take the limit, which you start to talk about next.
christian0710 said:
It must be the partial sum of the n-th term in the sequence an,
"the partial sum associated with the series." Leave off the other part.

IOW, we chop off the infinitely long tail of the infinite series and get a partial sum.
christian0710 said:
I remember from the text "The sum of a series is the limit of the sequence of partial sums"
YES! Now you're really getting there.
christian0710 said:
, so the sum of the series Ʃ1/2n must be the limit of Sn?
Yes, absolutely.
christian0710 said:
But does it make sense to speak about a limit of Sn = (2n - 1)/22 which is a finite number?
Yes, it sure does. Keep in mind that Sn is not a constant - its value depends on n. For example, S2 = 3/4, and S3 = 7/8.

Remember that S2 really means 1/2 + 1/4, and that S3 means 1/2 + 1/4 + 1/8. The formula above is just a compact way to write the general term in the sequence of partial sums.

As you suspect, what we do with Sn is take its limit and see if we get a value. If we do, the series we're investigating converges to (adds up to) that value. Conversely, if the limit doesn't exist, the series diverges.

For this simple example,
$$ \lim_{n \to \infty} \frac{2^n - 1}{2^n} = 1$$

This tells us that ## \sum_{i = 1}^{\infty} \frac{1}{2^n} = 1##
christian0710 said:
It's like saying "the limit of 25 = 25" right?
 
  • #23
It looks like we are getting to the core of this. There is still a thing about the partial sum Sn that confuses me:

I understand that: A given series an can go to infinity, but a partial sum is always a finite number, that never exceeds the n-th term in the series.

The confusion might be: The difference between Sn and {Sn}

Sn must be the partial sum of the series or of the sequence. So if we choose a fixed n in of the series, we get a fixed finite number when we choose a fixed n (So Sn is never infinity)

{Sn}={s1,s2,s3…sn…} This is the sequence which is formed from the partial sums (s1=a1, S2 =a1+a2 etc. ) and the (…) at the end signifies that the sequence of partial sums goes to infinity (It just seems strange that we let n go to infinity, but (…) shows that we exceed n which means we exceed infinity?), and if we let n go to infinity (which we can do in this case because we want to test for limits) we get the limit of the sequence of the partial sum, which is the sum of the series.

But how can we let n (or is it the the sequence {Sn}) go to infinity, if Sn cannot go to infinity (it’s a finite number)? Is it because, we found the the pattern in which Sn increases or decreases, so it’s like a function and to test the limit we imagine n goes to infinity?

jesus it’s 1.38Pm here in Demark. I got to go, can’t wait to learn more :)
 
  • #24
christian0710 said:
It looks like we are getting to the core of this. There is still a thing about the partial sum Sn that confuses me:

I understand that: A given series an can go to infinity, but a partial sum is always a finite number, that never exceeds the n-th term in the series.
A given series, abbreviated ##\sum a_n##, can diverge or converge.
For each n, a partial sum is always the sum of a finite number of terms, so always represents a number. Remember that we have chopped off all but a finite number of terms.
Sn is really a function whose domain is often the nonnegative integers or the positive integers. So the notation we're writing as Sn could just as well have been written as S(n).

What you wrote "... a partial sum is always a finite number, that never exceeds the n-th term in the series" is not generally true.

Think about things in terms of the series ## \sum_{n = 1}^{\infty}\frac{1}{2^n}## = 1/2 + 1/4 + 1/8 + ... 1/2n + ...

S1 = 1/2
S2 = 1/2 + 1/4 = 3/4
Note that S2 > a2, which is 1/4.

S3 = 1/2 + 1/4 + 1/8 = 7/8
Note that S3 > a3, which is 1/8.

And so on.
christian0710 said:
The confusion might be: The difference between Sn and {Sn}

Sn is the nth term in the sequence.
{Sn} is notation that represents all the terms in the sequence. IOW,
{Sn} means {S1, S2, S3, ..., Sn, ...}


christian0710 said:
Sn must be the partial sum of the series or of the sequence. So if we choose a fixed n in of the series, we get a fixed finite number when we choose a fixed n (So Sn is never infinity)

{Sn}={s1,s2,s3…sn…} This is the sequence which is formed from the partial sums (s1=a1, S2 =a1+a2 etc. ) and the (…) at the end signifies that the sequence of partial sums goes to infinity (It just seems strange that we let n go to infinity, but (…) shows that we exceed n which means we exceed infinity?)
The ... at the end means that the sequence continues in the same pattern. The next term in the sequence would be Sn + 1, and the one after that would be Sn + 2. That's all it means. It doesn't mean that the sequence of partial sums goes to infinity - it means there are an infinite number of terms in the sequence. There's a difference.

For example, if the sequence happened to be {1, 1, 1, 1, ..., 1, ...}, this represents an infinite number of terms, but the terms themselves aren't "going to infinity". This very simple sequence obviously converges to 1.
christian0710 said:
, and if we let n go to infinity (which we can do in this case because we want to test for limits) we get the limit of the sequence of the partial sum, which is the sum of the series.

But how can we let n (or is it the the sequence {Sn}) go to infinity,
We're letting n "get large". The sequence can do whatever it does - converge to a number, diverge to infinity, or flip-flop around, never settling down to a single value.

Some examples:
1. {1/2, 1/4, 1/8, ..., 1/2n, 1/2n+1, ...}
This sequence converges to 0. Obviously, for any finite value of n, an is always somewhat larger than zero, but the idea is that however close to zero we need to get, we can find a value of N so that from aN+1 on, all of these terms are smaller than that specified "closeness" to zero. That's really what we mean when we say that a sequence converges to some number. We never just plug in ∞ and see what we get (which would be meaningless to do).

2. {1, 2, 4, 8, ... , 2n, 2n + 1, ...}
This sequence diverges. As n gets larger, so also does an. This sequence diverges, or grows large without bound. What this means is that no matter how large a number someone says, we can find a number N so that aN+1 and all of the terms after it are larger than that specified value.

3. {1, 0, 1, 0, 1, 0, ...}
The general term in this sequence has two formulas: an = 1 if n is odd, and an = 0 if n is even.
This sequence diverges because it oscillates between 0 and 1.
christian0710 said:
if Sn cannot go to infinity (it’s a finite number)? Is it because, we found the the pattern in which Sn increases or decreases, so it’s like a function and to test the limit we imagine n goes to infinity?
Technically, we're letting n "increase without bound"
christian0710 said:
jesus it’s 1.38Pm here in Demark. I got to go, can’t wait to learn more :)
 
  • #25
I almost got it all now!

There is only ONE thing left that confuses me. It's the idea of taking the limit of a partial sum (Not the Sequence of a partial sum which makes more sense) which is what they do in my book.

The partial sum does not go to infinity according to the definition:
Sn=Ʃai=a1+a2+a3+...+an.
For me it makes sense to take the limit of a sequence of partial sums where the terms go to infinity (just like taking the limit of a function where x goes to infinity)

In my book under the phrase "The sum of the series is the limit of the sequence of the partial sums" they show
Ʃan=Lim Ʃai where Ʃai = Sn. How can they take a limit of a sum/series that only goes to n?

So i understand that we take the limit of a sequence Lim an= 1/2n we get 0
the sum of this series is 1, do we also say "the limit of the series" Lim Ʃ1/2n"?

This is the last part I'm having trouble understanding: taking limits af sequences (makes sense) vs. taking limits of sums/series (makes little sense to me)

I just can't grasp this; the limit 1/2n is 0 (makes sence), but the limit of Ʃ1/2n is what? "the limit of a sum" how Am i to imagine that?

Edit: I just added a picture the sum and limit of sum as I understand it, could this be the right way of understanding it?
 

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  • #26
christian0710 said:
I almost got it all now!

There is only ONE thing left that confuses me. It's the idea of taking the limit of a partial sum (Not the Sequence of a partial sum which makes more sense) which is what they do in my book.
They're taking the limit of a sequence, the sequence of partial sums.

I think I see where you are confused -- between the general form of the partial sum and a particular partial sum (for a specific value of n).

Maybe this will make it clearer. For the series, ##\sum_{i = 1}^{\infty}\frac{1}{2^n}##, the general term in the sequence of partial sums is ##\frac{2^n - 1}{2^n}##.

The second term in the sequence of partial sums is 3/4 (when n = 2). Since 3/4 is a constant, its limit as n gets large is just 3/4. For any specified value of n, the corresponding term in the sequence of partial sums is a constant; hence its limit is just that same number.

However, the general term in the sequence is ##\frac{2^n - 1}{2^n}##, and its limit as n gets large, is 1.

christian0710 said:
The partial sum does not go to infinity according to the definition:
Sn=Ʃai=a1+a2+a3+...+an.
Try to separate the ideas of n going to infinity versus the limit of the partial sum going to infinity as n goes to infinity. The three examples I had in my previous post showed sequences that had different behaviors as n "went to infinity."
christian0710 said:
For me it makes sense to take the limit of a sequence of partial sums where the terms go to infinity (just like taking the limit of a function where x goes to infinity)
...where n goes to infinity.

When we're talking about sequences (which are functions), the argument of the function is usually n, a discrete variable that takes integer values. A variable such as x is usually used to denote a continuous variable that can take on all of the real values in an interval.
christian0710 said:
In my book under the phrase "The sum of the series is the limit of the sequence of the partial sums" they show
Ʃan=Lim Ʃai where Ʃai = Sn. How can they take a limit of a sum/series that only goes to n?
You're omitting information about the index of the summation, so are obscuring some important information.

Sn = a1 + a2 + ... + an = ## \sum_{i = 1}^n a_i##.

The sum of the series (if such a sum exists) is
$$ \lim_{n \to \infty} S_n = \lim_{n \to \infty} \sum_{i = 1}^n a_i$$

Sn is a function of n. As n gets larger, Sn represents a (finite) sum with more terms.


christian0710 said:
So i understand that we take the limit of a sequence Lim an= 1/2n we get 0
the sum of this series is 1, do we also say "the limit of the series" Lim Ʃ1/2n"?
Remember what I said a lot earlier in this thread about a series being associated with two sequences.

For an arbitrary series ## \sum_{i = 1}^{\infty} a_i##, the two sequences are:
1. The sequence of terms in the series: {a1, a2, ..., an, ...}
2. The sequence of partial sums: {S1, S2, ..., Sn, ...}

For the series we have been discussing, where ai = 1/2i, we have
1. ## \lim_{i \to \infty} a_i = \lim_{i \to \infty} 1/2^i = 0##, and
2. ## \lim_{i \to \infty} S_i = \lim_{i \to \infty} \frac{2^i - 1}{2^i} = 1##

The limit that is most important is the second one - the limit of the sequence of partial sums. The first limit above, is of much less importance. For a series to converge, it is necessary for the limit of the terms in the series to be zero, but this is not sufficient.

For example, in the well-known harmonic series, ## \sum_{n= 1}^{\infty} \frac{1}{n}##,
## \lim_{n \to \infty} \frac{1}{n} = 0##, but it turns out that the series itself diverges. From this we can conclude that the limit of the sequence of partial sums must diverge.
christian0710 said:
This is the last part I'm having trouble understanding: taking limits af sequences (makes sense) vs. taking limits of sums/series (makes little sense to me)
You don't take the limit of a series (an infinite sum), but you can take the limit of a finite sum, where the number of terms in the sum depends on an index n (IOW, is a function of n).
[/quote]
christian0710 said:
I just can't grasp this; the limit 1/2n is 0 (makes sence), but the limit of Ʃ1/2n is what? "the limit of a sum" how Am i to imagine that?
See above.
christian0710 said:
Edit: I just added a picture the sum and limit of sum as I understand it, could this be the right way of understanding it?

The sum you show in the picture is a finite sum: 1 + 2 + 3 + ... + n.

The formula n(n + 1)/2 is the sum of the first n integers. It does not represent the sum of an infinite number of integers.

Here's a question you asked in the thing you attached:

And the limit of the formula n(n + 1)/2 is 1?

Does the sequence {1, 3, 6, 10, ...} look like it is approaching 1?
 
  • #27
However, the general term in the sequence is 2n−12n, and its limit as n gets large, is 1.
I see so we are interested in taking the limit of the general term in the sequence {Sn} and this limit is a constant (a y value on the xy axis if S is a function of n if such limit exists)

Sn is a function of n. As n gets larger, Sn represents a (finite) sum with more terms.
Yes This definition lim(Sn)=limn→∑ai as n--> ∞
I don't understand why they write this right under the text saying "the sum of the series is the limit of the sequence of partial sums" Unless Sn signifies the sequence of the partial sums? In that case S is a function of n and n goes to infinity when we take the limit (makes sense)

So is lim(Sn)=lim∑ai as n--> ∞ the same as Lim{Sn}

but you can take the limit of a finite sum, where the number of terms in the sum depends on an index n
So lim∑i=ai is the finite sum, and you can take the limit of this finite sum, so I hope it's the same of{Sn}

Darn I've been sitting all day trying to understand this, but that single equation lim(Sn)=limn→∑ai as n--> ∞ really bugs me. I understand everything else you write.
 
  • #28
YESSSS I finally get it!
The sequence of partial sums is {Sn} =S1,S2,S3,...,Sn=a1+a2+a3+...+an=Ʃai, so if we take the limit of Ʃai we are taking the limit of the sequence of partial sums,.


Because if i=1 we get
Ʃai=1= S1 = a1
Ʃai=2= S2= a1+a2
Ʃai=1=S3 = a1+a2+a3 and as n approaches infinity we get all the terms S1,S2,S3...Sn from The sequence of partial sums Ʃai (The Ʃai is like a generator or a machine gun that shoots out the sequences one by one, and by taking the limit of this sequence we get the sum of the series Ʃan which is the sum when adding up all the number in the sequence An as n approaches infinity.

(This is the difference between Ʃai and Ʃan:
Ʃai is the the sequence the partial sum, so it goes from i to n Because Sn goes from 1 to n)
Ʃan: is the series that goes from 1 to infinity.
 
  • #29
christian0710 said:
YESSSS I finally get it!
The sequence of partial sums is {Sn} =S1,S2,S3,...,Sn=a1+a2+a3+...+an=Ʃai, so if we take the limit of Ʃai we are taking the limit of the sequence of partial sums,.


Because if i=1 we get
Ʃai=1= S1 = a1
Ʃai=2= S2= a1+a2
Ʃai=1=S3 = a1+a2+a3 and as n approaches infinity we get all the terms S1,S2,S3...Sn from The sequence of partial sums Ʃai (The Ʃai is like a generator or a machine gun that shoots out the sequences one by one, and by taking the limit of this sequence we get the sum of the series Ʃan which is the sum when adding up all the number in the sequence An as n approaches infinity.

(This is the difference between Ʃai and Ʃan:
You're a little confused here.

## \sum_{i = 1}^n a_i## is a sum with n terms. This is exactly what I've been calling Sn.

Which letter you use for the index isn't very important, as long as you are consistent. All of these are the same as the above.
## \sum_{j = 1}^n a_j##
## \sum_{k= 1}^n a_k##
## \sum_{p = 1}^n a_p##


## \sum_{i = 1}^{\infty} a_i## is the (infinite) series.


christian0710 said:
Ʃai is the the sequence the partial sum, so it goes from i to n Because Sn goes from 1 to n)
Ʃan: is the series that goes from 1 to infinity.

You're omitting a lot of information in the two sentences above. Without additional information, there's no difference between Ʃai and Ʃan (don't write
Ʃan) unless you mean the series a1 + a2 + ... + an + ... This is a geometric series.
 
  • #30
You are probably right I'm confusing things again.

Sn= a1+a2+a3+...+an = The n'th partial sum (Not a sequence) = Ʃai (where i goes from 1 to n)

So this: LimƩai (where i goes from 1 to n) (the limit goes from n --> ∞)
How does that look? If the n'th partial sum only goes to n, but n goes to infinity, are we then finding the limit of How Sn (the n'th partial sum) would progress beyond n? So if the n'th partial sum has the pattern (n2-1)/22 then the limit of Sn=Ʃai is one? Just like taking the limit of a sequence?
 
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  • #31
christian0710 said:
You are probably right I'm confusing things again.

Sn= a1+a2+a3+...+an = The n'th partial sum (Not a sequence) = Ʃai (where i goes from 1 to n)

So this: LimƩai (where i goes from 1 to n) (the limit goes from n --> ∞)
How does that look? If the n'th partial sum only goes to n, but n goes to infinity, are we then finding the limit of How Sn (the n'th partial sum) would progress beyond n?
Not beyond n, which isn't a fixed number - how Sn progresses for larger values of n.
christian0710 said:
So if the n'th partial sum has the pattern (n2-1)/22 then the limit of Sn=Ʃai is one?
No, not even close. Try it yourself.

If n = 1, (n2 - 1)/4 = 0
If n = 2, (n2 - 1)/4 = 3/4
If n = 3, (n2 - 1)/4 = ?
Put in three or four more values for n and see what you get.
christian0710 said:
Just like taking the limit of a sequence?

The sequence of partial sums is a sequence, so yes.
 
  • #32
Not beyond n, which isn't a fixed number - how Sn progresses for larger values of n.

Okay, so maybe there is no confusion, So: LimƩai (where i goes from 1 to n) (the limit goes from n --> ∞) means How the partial sum Sn progresses as n goes to infinity So if we take this example:

Example
an = 1/2,1/4,1/8,,1/2n

{Sn} = 1/2, 3/4,7/8,...,(2n - 1)/2n
Sn= 1/2+ 1/4+1/8+...+ 1/2n
So LimƩai = Lim(Sn)=Lim(1/2+ 1/4+1/8+...+ 1/2n (where i goes from 1 to n) (the limit goes from n --> ∞) In my book they argue that THIS is the same as the sum of a series. Perhaps I should take a picture? because you see why it does not make sense? We are not taking the limit of the sequence of partial sums, we are taking the limit of the partial sum Sn. Ahh but wait, we are taking the limit of how the partial sum progresses, which is the same as the sequence of the partial sum (a progression)? If n = 1, (n2 - 1)/4 = 0
If n = 2, (n2 - 1)/4 = 3/4
If n = 3, (n2 - 1)/4 = 8/4 of yea this goes to infinity, my mistake :)
 
  • #33
christian0710 said:
Okay, so maybe there is no confusion, So: LimƩai (where i goes from 1 to n) (the limit goes from n --> ∞) means How the partial sum Sn progresses as n goes to infinity So if we take this example:

Example
an = 1/2,1/4,1/8,,1/2n

{Sn} = 1/2, 3/4,7/8,...,(2n - 1)/2n
Sn= 1/2+ 1/4+1/8+...+ 1/2n
So LimƩai = Lim(Sn)=Lim(1/2+ 1/4+1/8+...+ 1/2n (where i goes from 1 to n) (the limit goes from n --> ∞) In my book they argue that THIS is the same as the sum of a series. Perhaps I should take a picture? because you see why it does not make sense? We are not taking the limit of the sequence of partial sums, we are taking the limit of the partial sum Sn. Ahh but wait, we are taking the limit of how the partial sum progresses, which is the same as the sequence of the partial sum (a progression)?
... which is the same as the limit of the sequence of partial sums ...

If I have the formula for the general term in a sequence, it's much easier to find the limit of that sequence.

Knowing that Sn = 1/2 + 1/4 + 1/8 + ... + 1/2n doesn't do me much good if I want to find ## \lim_{n \to \infty} S_n##

But, if I also know that 1/2 + 1/4 + 1/8 + ... + 1/2n = ## \frac{2^n - 1}{2^n}## (which I can find by using induction or from knowledge about the sum of a finite geometric series), then I can find ## \lim_{n \to \infty} S_n## and, hence, the sum of the infinite series.

## \lim_{n \to \infty} S_n = \lim_{n \to \infty}\frac{2^n - 1}{2^n} = 1##

This allows me to say that 1/2 + 1/4 + 1/8 + ... + 1/2n + ... is a convergent series that adds up to (converges to) 1.


christian0710 said:
If n = 1, (n2 - 1)/4 = 0
If n = 2, (n2 - 1)/4 = 3/4
If n = 3, (n2 - 1)/4 = 8/4 of yea this goes to infinity, my mistake :)
 
  • #34
.. which is the same as the limit of the sequence of partial sums ...

YESS! :) That was what i actually meant, I finally understand it, thanks to your help!

But, if I also know that 1/2 + 1/4 + 1/8 + ... + 1/2n = (2n−1)/2n

So what you are doing here is actually finding the pattern for how the partial sum Sn progresses as n increases, which it does because we let n go to infinity and Lim (2n−1)/2n[/QUOTE] is the sum of the series :)

So in your case Lim Sn is the sum of the series and so Lim Sn must also be Lim{Sn] (the limit of the sequence of partial sums) Seems confusing, but by the way of argument i see how it can make sense.
 

1. What is an infinite sequence?

An infinite sequence is a list of numbers that continues on forever, with no end. Each number in the sequence is called a term, and the order of the terms is important.

2. What is the difference between a finite and an infinite sequence?

A finite sequence has a limited number of terms, while an infinite sequence has an unlimited number of terms. In other words, a finite sequence will eventually come to an end, while an infinite sequence will continue on forever.

3. How do you find the sum of an infinite series?

To find the sum of an infinite series, you can use a formula called the partial sum formula, which involves taking the limit of the partial sums of the series. Alternatively, you can also use various convergence tests to determine if the series converges and if so, what its sum is.

4. What is the difference between a convergent and a divergent series?

A convergent series is one in which the sum of all its terms approaches a finite number as the number of terms increases. In other words, the series "converges" to a specific value. A divergent series is one in which the sum of its terms either approaches infinity or does not approach any specific value.

5. How is the concept of infinite sequences and series used in real life?

Infinite sequences and series are used in various fields of science and mathematics, such as physics, engineering, and computer science. They are used to model and analyze real-life phenomena, such as the motion of objects, electrical circuits, and data compression. They also play a crucial role in the development of advanced mathematical concepts and theories.

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