# Find $S_n = n/n+1$, Find $A_n$, Series $\{A_n\}$ and $\lim A_n$

This is the problem that I am attempting to solve. I'm lost on it because I thought I had it, but I guess I didn't as the correct answer is shown below.

If the $$n$$ partial sum of a series $$\sum_{n=1}^\infty a_n$$ is $$s_n = \frac{n}{n+1}$$, find $$a_n$$, $$\sum_{n=1}^\infty a_n$$, and $$\lim\limits a_n$$.

This is the correct answer.

$$a_n = \frac{1}{n(n+1)}, \qquad \sum_{n=1}^\infty a_n =1, \qquad \lim\limits a_n=0$$

When originally solving this problem, I thought, well series $$\{A_n\} = \lim\limits_{n \to \infty} (S_n)$$ and $$S_n = n/(n+1)$$, thus series $$\{A_n\} = n/(n+1)$$, but this isn't the case in the answer.

Can anyone explain to me why not/direct me on the correct path to solving this problem?

Also, when I "solved it" I got that it diverges because of DIV test limits to 1, not 0, but that's wrong because my series An is wrong.

You have

$$s_n = \sum_{k=1}^n a_k$$

so that

$$s_{n}-s_{n-1} = \sum_{k=1}^n a_k - \sum_{k=1}^{n-1} a_k = a_n.$$

That is:

$$a_n = \frac{n}{n+1} - \frac{n-1}{n},$$

so that will give you the correct $$a_n$$.

Well a general fact about series in general:

If $$S_n \rightarrow c < \infty$$, then we can prove that for any series $$a_n$$ such that $$\sum a_n = S_n$$, then $$a_n \rightarrow 0$$.

This is powerful result that the problem asks you to compute for this specific series. The essence of this problem is asking you to decompose a partial sum into its components.

Another way to think about it is if you were infinitely summing something far bigger than zero, you couldn’t possibly converge to anything other than infinity. But $$S_n$$ does. So this forces the series in question to converge to zero at a “quick” rate. I say quick because the sequence $$\{ 1/n \}$$ doesn’t converge “fast” enough.