<s>[INST] 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?
 [/INST] An option 
$S_n:= \sum _{i=1}^{n}a_i  = $
$\dfrac{n-1}{n+1}=1- 2\dfrac{1}{n+1}.$
$\lim_{ n \rightarrow \infty} S_n = 1- 2\lim_{n \rightarrow \infty}\dfrac{1}{n+1}.$
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