Evaluating $\sum (-1)^{n+1} (n+1 + \frac{1}{n+1})/n!$ Let $a_{n}=n+\dfrac{1}{n}$ for $n \in \mathbb{N}$. Find the sum of series $$\sum_{n=1}^{\infty}(-1)^{n+1}\dfrac{a_{n+1}}{n!}.$$
This becomes:
$$\begin{align}
\sum_{n=1}^{\infty}(-1)^{n+1}\Big[\dfrac{(n+1)+\dfrac{1}{n+1}}{n!}\Big] &\implies
\sum_{n=1}^{\infty}(-1)^{n+1}\Big[\dfrac{(n+1)}{n!}+\dfrac{1}{(1+n)!}\Big] \\
&\implies \sum_{n=1}^{\infty}(-1)^{n+1}\dfrac{(n+1)}{n!}+\sum_{n=1}^{\infty}(-1)^{n+1}\dfrac{1}{(1+n)!}\\
& \implies \sum_{n=1}^{\infty}(-1)^{n+1}\dfrac{(n+1)}{n!}+e^{-1}-1
\end{align}$$
I don't know how to simplify further.
 A: Hint: $\displaystyle\frac{n+1}{n!}=\frac1{(n-1)!}+\frac1{n!}$.
A: $\begin{array}\\
\sum_{n=1}^{\infty} (-1)^{n+1} (n+1 + \frac{1}{n+1})/n!
&=\sum_{n=2}^{\infty} (-1)^{n} \dfrac{n + \frac{1}{n}}{(n-1)!}\\
&=\sum_{n=2}^{\infty} (-1)^{n} \dfrac{n^2 + 1}{n!}\\
&=\sum_{n=2}^{\infty} (-1)^{n} \dfrac{n^2}{n!}+\sum_{n=2}^{\infty} (-1)^{n} \dfrac{1}{n!}\\
&=\sum_{n=2}^{\infty} (-1)^{n} \dfrac{n(n-1)+n}{n!}+e^{-1}\\
&=\sum_{n=2}^{\infty} (-1)^{n} \dfrac{n(n-1)}{n!}+\sum_{n=2}^{\infty} (-1)^{n} \dfrac{n}{n!}+e^{-1}\\
&=\sum_{n=2}^{\infty} (-1)^{n} \dfrac{1}{(n-2)!}+\sum_{n=2}^{\infty} (-1)^{n} \dfrac{1}{(n-1)!}+e^{-1}\\
&=\sum_{n=0}^{\infty} (-1)^{n+2} \dfrac{1}{n!}+\sum_{n=1}^{\infty} (-1)^{n+1} \dfrac{1}{n!}+e^{-1}\\
&=1+\sum_{n=1}^{\infty} ((-1)^{n+2}+(-1)^{n+1}) \dfrac{1}{n!}+e^{-1}\\
&=1+e^{-1}\\
\end{array}
$
A: The sum is $\sum_{n\ge 0}(-1)^n(\frac{1}{n!}+\frac{1}{(n+1)!}+\frac{1}{(n+2)!})$. While $\frac{1}{0!},\,\frac{1}{1!}$ have respective coefficients $1,\,0$, any other $\frac{1}{n!}$ has coefficient $\sum_{k=n-2}^{n}(-1)^k=(-1)^n$. Thus $\frac{1}{1!}$ is the only term that doesn't get the same coefficient as in $e^{-1}$, scoring $0$ instead of $-1$. The final result is therefore $e^{-1}+1$.
