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.
 A: $$\text{From }\frac{n+1}{(n+2)!},$$
$$=\frac{n+2-1}{(n+2)!}=\frac1{(n+1)!}-\frac1{(n+2)!}$$
Can you identify the Telescoping series?
The survivor will be $$\frac1{2!}$$
A: Hint : $\displaystyle f(x)=\sum\frac{x^n}{(n+1)!}\iff f'(x)=\sum\frac{n\cdot x^{n-1}}{(n+1)!}\quad,\quad S=f'(1)$.
A: 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 $n=1$.)
A: $\dfrac{1}{3}+\dfrac{1}{4}\dfrac{1}{2!}+\dfrac{1}{5}\dfrac{1}{3!}+\dots\\=\displaystyle\sum_3^\infty\dfrac{1}{n}\times\dfrac{1}{(n-2)!}\\=\displaystyle\sum_3^\infty\dfrac{n-1}{n!}\\=\displaystyle\sum_3^\infty\left[\dfrac{1}{(n-1)!}-\dfrac{1}{n!}\right]\\=\dfrac{1}{2!}$
