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Infinite sequences and series.. help!

 Quote by christian0710 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 by christian0710 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 by christian0710 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 by christian0710 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 by christian0710 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 by christian0710 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 by christian0710 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 by christian0710 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!

 Great post Mark44!
 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?

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 Quote by jedishrfu Great post Mark44!
Thanks!

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 Quote by christian0710 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.
 Quote by christian0710 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.
 Quote by christian0710 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.
 Quote by christian0710 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.
 Quote by christian0710 , so the sum of the series Ʃ1/2n must be the limit of Sn?
Yes, absolutely.
 Quote by christian0710 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##
 Quote by christian0710 It's like saying "the limit of 25 = 25" right?

 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 :)

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 Quote by 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.
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.
 Quote by christian0710 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, ...}

 Quote by christian0710 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.
 Quote by christian0710 , 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.
 Quote by christian0710 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"
 Quote by christian0710 jesus it’s 1.38Pm here in Demark. I got to go, can’t wait to learn more :)

 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? Attached Thumbnails

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 Quote by 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.
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.

 Quote by christian0710 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."
 Quote by christian0710 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.
 Quote by christian0710 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.

 Quote by christian0710 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.
 Quote by christian0710 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).
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 Quote by christian0710 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.
 Quote by christian0710 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?

 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.

 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.

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 Quote by christian0710 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.

 Quote by christian0710 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.

 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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 Quote by christian0710 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.
 Quote by christian0710 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.
 Quote by christian0710 Just like taking the limit of a sequence?
The sequence of partial sums is a sequence, so yes.

 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 :)

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 Quote by christian0710 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.

 Quote by christian0710 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 :)

 .. 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.