Sum of series 5.



\(\displaystyle \;\; \sum_{n=0}^{\infty} \frac{1}{(n+2) n!}\)

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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