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.
Best Answer
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 $a_n$.