Homework Help: Infinite sequences and series help!

1. Dec 7, 2012

christian0710

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. Dec 7, 2012

Staff: Mentor

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. Dec 7, 2012

christian0710

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. Dec 7, 2012

christian0710

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. Dec 7, 2012

christian0710

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. Dec 7, 2012

Staff: Mentor

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. Dec 7, 2012

chief10

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. Dec 7, 2012

christian0710

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

Last edited: Dec 7, 2012
9. Dec 7, 2012

Staff: Mentor

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...

With respect to this statement it appears to be correct.

10. Dec 7, 2012

christian0710

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. Dec 7, 2012

Staff: Mentor

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.

12. Dec 7, 2012

christian0710

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. Dec 7, 2012

christian0710

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. Dec 7, 2012

christian0710

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. Dec 7, 2012

Staff: Mentor

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.

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. Dec 7, 2012

christian0710

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.

Last edited: Dec 7, 2012
17. Dec 7, 2012

christian0710

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. Dec 7, 2012

Staff: Mentor

No need for apology. This is complicated stuff when you're first exposed to 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.
If the series is convergent, then it represents a number. Other series are divergent, so they don't represent a number.
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.
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.
The last term that you show should be 1/2n.
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, ...}
We say "converges to 1", not "toward".

Good work!

19. Dec 7, 2012

Staff: Mentor

Great post Mark44!

20. Dec 7, 2012

christian0710

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?

21. Dec 7, 2012

Staff: Mentor

Thanks!

22. Dec 7, 2012

Staff: Mentor

Corrected a couple of your subscripts that should be exponents in the following.
It gives you a handy, dandy form that is ripe for you to take the limit, which you start to talk about next.
"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.
YES! Now you're really getting there.
Yes, absolutely.
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 $\sum_{i = 1}^{\infty} \frac{1}{2^n} = 1$

23. Dec 7, 2012

christian0710

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. Dec 8, 2012

Staff: Mentor

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.
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, ...}

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.
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.
Technically, we're letting n "increase without bound"

25. Dec 8, 2012

christian0710

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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