Infinite sequences and series.. help!

Mark44

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.
S_n represents the sum of a finite number of terms for each value of n.
S_n = a₁ + a₂ + ... + a_n. However, as n increases, the number of terms that make up S_n 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!

jedishrfu

Great post Mark44!

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

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

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 = (2_n - 1)/2² which is a finite number? It's like saying "the limit of 25 = 25" right?

Mark44

Thanks!

Mark44

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 S_n 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 S_n 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

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 a_n 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 :)

Mark44

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.
S_n is really a function whose domain is often the nonnegative integers or the positive integers. So the notation we're writing as S_n 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.
S_n is the nth term in the sequence.
{S_n} is notation that represents all the terms in the sequence. IOW,
{S_n} means {S₁, S₂, S₃, ..., S_n, ...}


The ... at the end means that the sequence continues in the same pattern. The next term in the sequence would be S_{n + 1}, and the one after that would be S_{n + 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, a_n 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 a_N+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, 2^{n + 1}, ...}
This sequence diverges. As n gets larger, so also does a_n. 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 a_N+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: a_n = 1 if n is odd, and a_n = 0 if n is even.
This sequence diverges because it oscillates between 0 and 1.
Technically, we're letting n "increase without bound"

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
Ʃa_n=Lim Ʃa_i where Ʃa_i = 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

 

Mark44

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.

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."
...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.
You're omitting information about the index of the summation, so are obscuring some important information.

S_n = a₁ + a₂ + ... + a_n = ## \sum_{i = 1}^n a_i##.

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

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


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, ..., a_n, ...}
2. The sequence of partial sums: {S₁, S₂, ..., S_n, ...}

For the series we have been discussing, where a_i = 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.
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]
See above.
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:

Does the sequence {1, 3, 6, 10, ...} look like it is approaching 1?

christian0710

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)

Yes This definition lim(Sn)=limn→∑a_i 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∑a_i as n--> ∞ the same as Lim{Sn}

So lim∑i=a_i 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→∑a_i as n--> ∞ really bugs me. I understand everything else you write.

christian0710

YESSSS I finally get it!!!
The sequence of partial sums is {Sn} =S1,S2,S3,...,Sn=a1+a2+a3+...+an=Ʃa_i, so if we take the limit of Ʃa_i 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 Ʃa_i (The Ʃa_i 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 Ʃa_i and Ʃaⁿ:
Ʃa_i 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.

Mark44

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 S_n.

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.


You're omitting a lot of information in the two sentences above. Without additional information, there's no difference between Ʃa_i and Ʃa_n (don't write
Ʃaⁿ) unless you mean the series a¹ + a² + ... + aⁿ + ... This is a geometric series.

christian0710

You are probably right I'm confusing things again.

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

So this: LimƩa_i (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=Ʃa_i is one? Just like taking the limit of a sequence?

Mark44

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

christian0710

Okay, so maybe there is no confusion, So: LimƩa_i (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
a_n = 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 LimƩa_i = 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 :)

Mark44

... 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 S_n = 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

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

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.