# If the $n^{th}$ partial sum of a series $\sum a_n$ is $S_n = \frac{n-1}{n+1}$, find $a_n$ and the sum.

If the $n^{th}$ partial sum of a series $\sum a_n$ is $S_n = \dfrac{n-1}{n+1}$, find $a_n$ and the sum.

By definition, the sum of the series is the $\lim n\to\infty$ of it’s $n^{th}$ partial sum.

$$\text{ Sum = } \lim_{n\to\infty} = \dfrac{n-1}{n+1} = 1$$

I am asked to find $a_n$. How do I do this? What is the procedure?

#### Solutions Collecting From Web of "If the $n^{th}$ partial sum of a series $\sum a_n$ is $S_n = \frac{n-1}{n+1}$, find $a_n$ and the sum."

Let $S_n$ be the partial sum of a sequence $a_n$. Then, we can write

$$S_n=\sum_{k=1}^na_k \tag1$$

for $n\ge 1$.

Next, using the hint in the comment from S.B. Art, we have from $(1)$

$$a_n = S_{n}-S_{n-1}=\frac{2}{n}-\frac2{n+1} \tag2$$

for $n\ge 2$. The expression in $(2)$ is not valid for $n=1$ since $S_0$ is not defined by $(1)$.

However, we are given $S_n=\frac{n-1}{n+1}$. Thus, $S_1=0$ and hence $a_1=0$ also.

Finally, we can write

$$a_k=\begin{cases}\frac2k-\frac2{k+1}&,k\ge 2\\\\0&,k=1\end{cases}$$