Infinite sequences and series help

[Mark44]

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?

Not beyond n, which isn't a fixed number - how S_n progresses for larger values of n.

So if the n'th partial sum has the pattern (n²-1)/2² then the limit of Sn=Ʃa_i 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.

Just like taking the limit of a sequence?

The sequence of partial sums is a sequence, so yes.

 

[christian0710]

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: 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]

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)?

... 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.

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 :)

 

[christian0710]

.. 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.