Why Doesn't S_n Converging to 1 Mean the Series Diverges?

[FancyNut]

I'm reading How to Ace the Rest of Calculus and on page 31 there's a test for divergence that says if the limit (as n goes to infinity) of a_n is NOT equal to zero, then the infinite series \sum a_n (that goes from n = 1 to infinity) diverges.

3 pages before that, on 28, there's an example on a series that converges to 1... The sum is represented by S_n = (2^n - 1)/2^n

I'm confused here: since taking the limit of that S_n equals to 1... shouldn't the series diverge? Come to think of it I know I'm missing something here because as I understand it according to that test ANY series that has a limit to any number but 0 diverges which makes no sense. I know that test does nor work in reverse btw (limit going to 0 doesn't mean it converges) but this still sounds like all series convergence must go to 0... which I know is wrong.

 

[arildno]

You are confusing a_{n} with S_{n}
a_{n} is just some sequence of numbers, while S_{n} is the sum of the first n a_{j} (j=1,2...n)
Was that clear enough?

 

[shmoe]

With the divergence test you look at the limit of the terms in your series, the a_n's. This is different from the limit of the partial sums, the S_n's. Note S_n=\sum_{i=1}^{n}a_i and we define \sum_{i=1}^{\infty}a_i=\lim_{n\rightarrow\infty}S_n

The example you gave, the partial sums S_n = (2^n - 1)/2^n, came from the terms a_n=1/2^n. Here \lim_{n\rightarrow\infty}a_n=0 so the divergence test is inconclusive. The partial sums converge to 1, so we say the series converges.

 

[FancyNut]

haha that was simple.

Thanks.