What is $\lim_{n\to\infty}\sum_{r=1}^{n}\frac{1}{(r+2)r!}\:$? $$\lim_{n\to\infty}\displaystyle\sum_{r=1}^{n}\frac{1}{(r+2)r!}$$
if we can find sum of this series then we can evaluate limit
\begin{equation*}
\frac{1}{(3)1!}+ \frac{1}{(4)2!}+\frac{1}{(5)3!}+ .... \frac{1}{(n+2)n!}
\end{equation*}
How we can find sum of this series 

First find $\displaystyle\sum_{r=1}^{n}\frac{1}{(r+2)r!}$ for this let 
Tr( rth term ) = $\frac{1}{(r+2)r!}$ or tn(nth term ) = $\frac{1}{(n+2)n!}$
Now simplyfying tn ..
$\frac{n+1}{(n+2)(n+1)n!}$
$\frac{n+1}{(n+2)!}$
now let $n+1=a_1+a_2(n+2) $and find value of $a_1$ and $a_2 $that is $a_1$=-1 and $a_2 $=1  , 
we get tn = $\frac{a_1+a_2(n+2)}{(n+2)!}$
or 
tn=$\frac{1}{(n+2)!}-\frac{1}{(n+1)!}$
put value of n=1,2,3,4,.... and add all term you will get sum =$\frac{1}{2}$
now , limit can we evaluated easily that is $\frac{1}{2}$
