<s>[INST] Evaluate $\frac{1}{3}+\frac{1}{4}\frac{1}{2!}+\frac{1}{5}\frac{1}{3!}+\dots$ Question is to Evaluate : 
$$\frac{1}{3}+\frac{1}{4}\frac{1}{2!}+\frac{1}{5}\frac{1}{3!}+\dots$$
what all i could do is :
$$\frac{1}{3}+\frac{1}{4}\frac{1}{2!}+\frac{1}{5}\frac{1}{3!}+\dots=\sum_{n=1}^{\infty} \frac{1}{(n+2)n!}=\sum_{n=1}^{\infty} \frac{n+1}{(n+2)!}=\sum_{n=1}^{\infty} \frac{n}{(n+2)!}+\sum_{n=1}^{\infty} \frac{1}{(n+2)!}$$
I have $$\sum_{n=1}^{\infty} \frac{1}{(n+2)!}=\sum_{n=0}^{\infty} \frac{1}{n!}-1-\frac{1}{2}=e-\frac{3}{2}$$
Now, I am not able to see what $$\sum_{n=1}^{\infty} \frac{n}{(n+2)!}$$ would be.
I would be thankful if some one can help me to clear this.
Thank you.
 [/INST] Since $\displaystyle e^x=\sum_{n=0}^\infty \frac {x^n}{n!}$, then $\displaystyle e^x-1-x=\sum_{n=0}^\infty\frac{x^{n+2}}{(n+2)!}$, and 
 $$ \frac{e^x-1-x}{x}=\sum_{n=0}^\infty\frac{x^{n+1}}{(n+2)!}, $$
so 
 $$ \left(\frac{e^x-1-x}{x}\right)'=\sum_{n=0}^\infty\frac{(n+1)x^n}{(n+2)!}. $$
Evaluate at $x=1$, and subtract $1/2$ (the term corresponding to $n=0$) to find the value you are after. You obtain
 $$ \frac12=1-\frac12=\left(\frac{e^x(x-1)+1}{x^2}\right)_{x=1}-\frac12=\sum_{n=1}^\infty\frac{n+1}{(n+2)!}. $$
(By the way, you have a small typo in your analysis, since $\displaystyle e=\sum_{n=0}^\infty\frac1{n!}$; on line $6$ you wrote the series starting at $n=1$.)
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