<s>[INST] Partial sum back into series. If the $n^{th}$ partial sum of a series $\sum a_n$ is $S_n = \frac{n−1}{n+1}$
then find $a_n$ and find
the sum of this series. Fully justify your answer.
Using the definition.
$a_n = S_{n} - S_{n-1} $
$$a_n = \frac{n-1}{n+1} - \frac{n-2}{n}$$
$$= \frac{n(n-1) - (n-2)(n+1)}{n(n+1)}$$
$$= \frac{n^2-n - (n^2-n-2)}{n(n+1)}$$
$$= \frac{2}{n(n+1)}$$
Hmm, I can use PFD on this to make it. 
$$= \frac{2}{n} - \frac{2}{n+1}$$
Is this right? What about n = 0 
The sum is 1 right? cus lim at $\infty$ ? 
 [/INST] Obviously, the sequence can be defined as
$$\begin{cases}a_0=S_0=-1,\\a_n=S_n-S_{n-1}=\dfrac{n-1}{n+1}-\dfrac{n-2}{n}=\dfrac2{n(n+1)},n>0.\end{cases}$$
(In particular, $a_1=1$, ensuring $S_1=a_0+a_1=0$.)
The sum is indeed $\lim_{n\to\infty}\dfrac{n+1-2}{n+1}=1$.
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