Infinite sequences and series.. help!


by christian0710
Tags: infinite, sequences, series
christian0710
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#1
Dec7-12, 06:09 AM
P: 156
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?
Attached Thumbnails
sum..PNG  
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jedishrfu
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#2
Dec7-12, 06:23 AM
P: 2,475
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
christian0710
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#3
Dec7-12, 06:45 AM
P: 156
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?
Attached Thumbnails
sum.JPG  

christian0710
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#4
Dec7-12, 06:52 AM
P: 156

Infinite sequences and series.. help!


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 :)
christian0710
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#5
Dec7-12, 07:00 AM
P: 156
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?
jedishrfu
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#6
Dec7-12, 08:02 AM
P: 2,475
Quote Quote by christian0710 View Post
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 :)
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
chief10
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#7
Dec7-12, 10:07 AM
P: 78
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
christian0710
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#8
Dec7-12, 11:17 AM
P: 156
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
jedishrfu
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#9
Dec7-12, 11:30 AM
P: 2,475
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...



Quote Quote by christian0710 View Post
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.
christian0710
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#10
Dec7-12, 11:39 AM
P: 156
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.
Mark44
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#11
Dec7-12, 12:51 PM
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Quote Quote by christian0710 View Post
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.
Quote Quote by christian0710 View Post
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.
christian0710
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#12
Dec7-12, 01:09 PM
P: 156
Quote Quote by Mark44 View Post
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.
christian0710
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#13
Dec7-12, 01:13 PM
P: 156
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?
christian0710
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#14
Dec7-12, 01:19 PM
P: 156
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"
Mark44
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#15
Dec7-12, 01:46 PM
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Quote Quote by christian0710 View Post
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.

Quote Quote by christian0710 View Post
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.
christian0710
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#16
Dec7-12, 02:11 PM
P: 156
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.
christian0710
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#17
Dec7-12, 02:22 PM
P: 156
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! :)
Mark44
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#18
Dec7-12, 03:23 PM
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Quote Quote by christian0710 View Post
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.
Quote Quote by christian0710 View Post
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.
Quote Quote by christian0710 View Post

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.
Quote Quote by christian0710 View Post

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.
Quote Quote by christian0710 View Post
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 in to a particular value. Hence ##\sum_{i = 1}^{\infty} (-1)^n## is divergent.
Quote Quote by christian0710 View Post

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.
Quote Quote by christian0710 View Post

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, ...}
Quote Quote by christian0710 View Post
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!


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