<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] The trick is to find a useful simplification of $a_n$. This takes a certain amount of inspiration. It’s not obvious from the first few values; you might spot it from the first few values of $1+a_n$, but then again you might not. It’s tempting to rewrite $k^2+k+1$ as $\frac{k^3-1}{k-1}$, but that doesn’t immediately suggest any further simplification, so you might try to do something directly with $k!(k^2+k+1)$. Since $k^2+k=k(k+1)$, and the factor of $k+1$ combines nicely with $k!$, you might try that:
$$k!(k^2+k+1)=k(k+1)k!+k!=k(k+1)!+k!=k\Big((k+1)!+(k-1)!\Big)\;.$$
Then
$$\begin{align*}
a_n&=\sum_{k=1}^nk\Big((k+1)!+(k-1)!\Big)\\\\
&=\sum_{k=1}^nk(k-1)!+\sum_{k=1}^nk(k+1)!\\\\
&=\sum_{k=0}^{n-1}(k+1)k!+\sum_{k=2}^{n+1}(k-1)k!\\\\
&=1+2+\sum_{k=2}^{n-1}(k+1)k!+\sum_{k=2}^{n-1}(k-1)k!+(n-1)n!+n(n+1)!\\\\
&=3+\sum_{k=2}^{n-1}\Big((k+1)+(k-1)\Big)k!+(n-1)n!+n(n+1)!\\\\
&=3+2\sum_{k=2}^{n-1}kk!+(n-1)n!+n(n+1)!\\\\
&=3+2\sum_{k=2}^{n-1}\Big((k+1)k!-k!\Big)+(n-1)n!+n(n+1)!\\\\
&=3+2\sum_{k=2}^{n-1}\Big((k+1)!-k!\Big)+(n-1)n!+n(n+1)!\\\\
&=3+2(n!-2!)+(n-1)n!+n(n+1)!\\\\
&=(n+1)n!+n(n+1)!-1\\\\
&=(n+1)(n+1)!-1\;,
\end{align*}$$
so $1+a_n=(n+1)(n+1)!$. Now get a closed form for $\sum_{k=1}^n2k$, simplify, and write the series as a telescoping series.
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