SOLUTION : Multiply \(\;(n+1)\;\) in denominator and numerator.
\(\displaystyle \sum_{n=0}^{\infty} \frac{(n+1)}{(n+1)(n+2) n!}\)
\(\displaystyle=\sum_{n=0}^{\infty} \frac{(n+1)}{(n+1)(n+2)n!}\)
\(\displaystyle\Rightarrow \sum_{n=0}^{\infty} \frac{n}{(n+2) !}+\sum_{n=0}^{\infty} \frac{1}{(n+2) !}\)
\(\displaystyle=\sum_{n=0}^{\infty}\left[\frac{1}{(n+1)!}-\frac{2}{(n+2) !}\right]+\sum_{n=0}^{\infty} \frac{1}{(n+2) !} \)
\(\displaystyle=\sum_{n=0}^{\infty} \frac{1}{(n+1) !}-2 \sum_{n=0}^{\infty} \frac{1}{(n+2) !}+\sum_{n=0}^{\infty} \frac{1}{
(n+2)!} \;\; \cdots (1)\)
We know that \(\;\;\displaystyle \sum_{n=0}^{\infty} \frac{1}{(n)!}=e\)
We are going to use above series to solve the (1)
\(\displaystyle \sum_{n=0}^{\infty} \frac{1}{(n+1) !} -\sum_{n=0}^{\infty} \frac{1}{(n+2)!}\)
\(\displaystyle [e-1]-\left[e-2 \right]\)
\(\displaystyle \Rightarrow e-1-e+2=1\)
ANSWER : \(\displaystyle 1\)
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