Find the sum of the series? I need to find the sum of this series, where k is from 1 to infinity? 
$$\sum_{k=1}^\infty\frac{k^3+6k^2+11k+5}{(k+3)!}$$

HINT:
As the numerator is cubic, we need to express  $\displaystyle k^3+6k^2+11k+5$
as $1\cdot(k+3)(k+2)(k+1)+b_1(k+3)(k+2)+b_2(k+3)+b_3$ where $b_i$s are arbitrary   constants 
Observe  that we have to start with $k+3$ as it is there in the factorial of the denominator.
Had the numerator been square polynomial like $A k^2+Bk+C$ with the denominator $(k+r)!$
we had to start with $A\cdot (k+r)(k+r-1)+a_1(k+r)+a_2$ 

Here we have
$$\frac{k^3+6k^2+11k+5}{(k+3)!}=\frac{(k+3)(k+2)(k+1)+b_1(k+3)(k+2)+b_2(k+3)+b_3}{(k+3)!}$$
$$=\frac1{(k)!}+b_1\frac1{(k+1)!}+b_2\frac1{(k+2)!}+b_3\frac1{(k+3)!}$$
Multiplying either sides by $(k+3)!$ we get $$k^3+6k^2+11k+5=(k+3)(k+2)(k+1)+b_1(k+3)(k+2)+b_2(k+3)+b_3$$
Methods to determine $b_i$s:
Method $\#1:$
Set $k+3=0$ to find $b_3$
$k+2=0$ to find $b_2$ in terms of $b_3$ which is known
and so on
Method $\#2:$
Compare the coefficients of the different powers of $k$
