Sum of series 1.

\(\;\;\displaystyle \sum_{n=1}^{\infty}(-1)^{n+1} \frac{(n+1)^{2}+1}{(n+1) n !}\)

SOLUTION :

\(\displaystyle\sum_{n=1}^{\infty}(-1)^{n+1}\left[\frac{(n+1)^{2}}{(n+1) n !}+\frac{1}{(n+1) n !}\right]\)

\(\displaystyle\sum_{n=1}^{\infty}(-1)^{n+1}\left[\frac{n+1}{n !}+\frac{1}{(n+1) n !}\right]\)

\(\displaystyle\sum_{n=1}^{\infty}(-1)^{n+1}\left[\frac{n}{n !}+\frac{1}{n !}+\frac{1}{(n+1) !}\right]\)

\(\displaystyle\sum_{n=1}^{\infty}(-1)^{n+1}\left[\frac{1}{(n-1) !}+\frac{1}{n !}+\frac{1}{(n+1) !}\right]\)

\(\displaystyle\sum_{n=1}^{\infty} \frac{(-1)^{n+1}}{(n-1) !}+\sum_{n=1}^{\infty} \frac{(-1)^{n+1}}{n !}+\sum_{n=1}^{\infty} \frac{(-1)^{n+1}}{(n+1) !}\)

\(\displaystyle\left[1-1+\frac{1}{2 !}-\frac{1}{3 !}\cdots \right]+\left[1-\frac{1}{2 !}+\frac{1}{3 !}+\cdots\right]+\left[\frac{1}{2 !}-\frac{1}{3!}+\frac{1}{4 !}+\cdots\right]\)

\(\displaystyle e^{-1}-\left[-1+\frac{1}{2 !}-\frac{1}{3 !}+\cdots \cdot\right]+e^{-1}\)

\(\displaystyle\Rightarrow e^{-1}-\left[e^{-1}-1\right]+e^{-1}\)

\(\displaystyle\Rightarrow e^{-1}-e^{-1}+1+e^{-1}\)

\(\displaystyle\Rightarrow \quad 1+e^{-1}\)

\(\displaystyle\Rightarrow 1+\frac{1}{e}\)

ANSWER : \(\displaystyle 1+\frac{1}{e}\)