# Infinite series: nth-term test

1. Oct 22, 2004

### 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.

2. Oct 22, 2004

### 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?

3. Oct 22, 2004

### 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.

4. Oct 22, 2004

### FancyNut

haha that was simple.

Thanks.