Evaluate $\sum\limits_{i=1}^\infty a_i $ when $\sum\limits_{i=1}^na_i=\frac {n-1}{n+1}$ Question: $S_n=\frac {n-1}{n+1}$; Find $\sum_1^\infty a_i $
My answer: $a_n=S_n -S_{n-1}=\frac{2}{n^2+n}=\frac{2}{n}-\frac{2}{n+1}$
$\sum_1^\infty a_i=(\frac{2}{1}-\frac{2}{2})+(\frac{2}{2}-\frac{2}{3})+(\frac{2}{3}-\frac{2}{4})+......(\frac{2}{n}-\frac{2}{n+1}) = 2-\frac{2}{n+1}=2$
Am I correct in computing $a_n$ first and then finding $\sum_1^\infty a_i$? Or should I have done $\lim\limits_{n \to \infty}S_n =\lim\limits_{n \to \infty}\frac{n-1}{n+1}=1.$ Which is the correct method of evaluating this?
 A: As I know, the value of an infinite sum is the limit of its partial sums, in this case:
$$\sum_{i=1}^\infty a_i=\lim_{n\to\infty}S_n=1. $$
A: It's fine, except that you have a mistake in the calculation. Indeed the equation for $a_n$ isn't valid for $a_1$. You should have $a_1 = S_1 = 0$ and so the sequence starts from $a_2$, not $a_1$. In this way both results will coincide
Anyway what you do in the first step is finding $a_i$ explicitly and afterwards you use them to get $S_n$. Indeed there's no need for it and you can just use the fact that $\sum_1^{\infty} a_i = \lim_{n \to \infty} S_n$
A: By definition, the sum of a series is the limit of the sequence of its partial sums.
So, your second solution (sum = 1) is the correct one.
Remember that a_1 = S_1, so that a_1 = \frac{1-1}{1+1} = 0 (and not 1, as you had it).
A: Indeed, it is a general fact that $$\sum_{i=1}^{\infty} a_i = \lim_{n \to \infty} \sum_{i=1}^{n}a_i.$$ For this case, $$\lim_{n \to \infty} \sum_{i=1}^{n}a_i=\lim_{n \to \infty}\frac{n-1}{n+1}=\lim_{n\to \infty}\frac{1-\frac{1}{n}}{1+\frac{1}{n}}=1.$$ 
A: An option 
$S_n:= \sum _{i=1}^{n}a_i  = $
$\dfrac{n-1}{n+1}=1- 2\dfrac{1}{n+1}.$
$\lim_{ n \rightarrow \infty} S_n = 1- 2\lim_{n \rightarrow \infty}\dfrac{1}{n+1}.$
