If $a_n = n +{1 \over n}$ then find $$\sum_{n=1}^{\infty}{(-1)^{n+1}}{a_{n} \over n!}$$
My work :
$e^{-x}=1-\frac{x}{1!}+\cdots$ $$\sum (-1)^{n+1}\left(n+{1 \over n}\right)\cdot{1 \over n!}$$ $$=e^{-1}-\sum(-1)^{n+1}{1 \over {n\cdot n!}}$$ Now how can I proceed further? Thanks in advance. I am new here feel free to edit.
By using the convergent serie of the exponential integral (https://en.wikipedia.org/wiki/Exponential_integral#Convergent_series)
and $-E_{1}(x)=\text{Ei}(-x)$ for positive $x$ we get,
$$ \sum^{\infty}_{n=1}\frac{(-1)^{n+1}}{n\cdot n!} = -\sum^{\infty}_{n=1}\frac{(-1)^{n}}{n\cdot n!} =E_1(1)+\gamma=\gamma-\text{Ei}(-1) $$ so $$ \sum_{n=1}^{\infty}{(-1)^{n+1}}{a_{n} \over n!}=e^{-1}+\sum(-1)^{n+1}{1 \over {n\cdot n!}}= e^{-1}+\gamma-\text{Ei}(-1) $$