<s>[INST] Find $\sum ^{\infty }_{n=2}\dfrac {2+4+6+\ldots +2n}{1+a_{n}}$ Define $a_{n}=1!\left( 1^{1}+1+1\right) +2!\left( 2^{2}+2+1\right) +3!\left( 3^{2}+3+1\right) +\ldots +n!\left( n^{2}+n+1\right) $ , Find $\sum ^{\infty }_{n=2}\dfrac {2+4+6+\ldots +2n}{1+a_{n}}$
 [/INST] Observe that the $r$th term $(1\le r\le n)$ of $a_n$ is $\displaystyle(r^2+r+1)\cdot r!$
Now as $r^2+r+1$ is Quadratic, we can write  $\displaystyle r^2+r+1=(r+2)(r+1)+A(r+1)+B$ where $A,B$ are arbitrary constants, so that
$\displaystyle (r^2+r+1)\cdot r!=\{(r+2)(r+1)+A(r+1)+B\}\cdot r!$
$\displaystyle=(r+2)!+A\cdot (r+1)! +B\cdot r!$
If it were a cubic polynomial, we had to take $\displaystyle(r+3)(r+2)(r+1)+A(r+2)(r+1)+B(r+1)+C$ and so on.
Anyway, let us determine $A,B:$
we have  $\displaystyle r^2+r+1=(r+2)(r+1)+A(r+1)+B$
$\displaystyle\implies r^2+r+1=r^2+r(3+A)+2+A+B$
Comparing the coefficients of $\displaystyle r,3+A=1\implies A=-2$
Comparing the constants $\displaystyle 2+A+B=1\implies B=1$
$\displaystyle\implies r!(r^2+r+1)=(r+2)!- 2(r+1)!+r!$
$\displaystyle=\underbrace{(r+2)!- (r+1)!}-\underbrace{\{(r+1)!-r!\}}$
Observe the Telescoping Series 
Set $\displaystyle r=1,2,\cdots,n$ to find 
$\displaystyle a_n=(n+2)!-2!-\{(n+1)!-1\}=(n+1)!(n+2-1)-1$
$\displaystyle=(n+1)\cdot(n+1)!-1$
As Cocopuffs has derived each summand to be $\displaystyle=\frac{n(n+1)}{(n+1)\cdot(n+1)!}=\frac n{(n+1)!}$
$$\text{But, }\frac n{(n+1)!}=\frac {n+1-1}{(n+1)!}=\frac1{n!}-\frac1{(n+1)!}$$ 
Again, another Telescoping Series
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