Sum of series in the form $\sum^{\infty}_{n=0}\frac{1}{(n+k)\cdot n!}$ for some k>0 I was recently given the series $$\sum^{\infty}_{n=0}\frac{1}{(n+3)\cdot n!}$$ and was told that it evaluated to a very nice number of $e-2$.
However, I do not know how to arrive at that answer using any of the tricks I know. It is neither a geometric series, a p series, or a known Taylor series. Neither can I find any way to form an integral representation of $\frac{1}{(n+3)\cdot n!}$ so I can swap the integral and summation.
So how would I go about evaluating this?
More generally, how would I evaluate series with the form $$\sum^{\infty}_{n=0}\frac{1}{(n+k)\cdot n!}$$ for a positive nonzero k?
 A: Taylor series should be work well, since we need to find $S=\displaystyle \sum_{n\geqslant 0}\frac{1}{n!(n+3)}$ so define $\displaystyle f(x):=\sum_{n\geqslant 0}\frac{x^{n+3}}{n!(n+3)}$, then $f'(x)=x^{2}e^{x}$ so $\displaystyle f(x)=\int_{0}^{x}t^{2}e^{t}\, {\rm d}t$ and therefore $S=f(1)$. The same method work for the general case $\displaystyle \sum_{n\geqslant 0}\frac{1}{n!(n+k)}=\int_{0}^{1}x^{k-1}e^{x}\, {\rm d}x=(-1)^{-k}\Gamma(k,0,-1)$ and $k>0$ according Mathematica.
A: I think the nice comment of @Svyatoslav deserves an answer by its own.
We recall the identity
\begin{align*}
\int_{0}^1x^n\,dx=\frac{1}{n+1}
\end{align*}
and obtain
\begin{align*}
\color{blue}{\sum_{n=0}^\infty\frac{1}{(n+k)n!}}&=\sum_{n=0}^\infty\int_0^1x^{n+k-1}\,dx\frac{1}{n!}\\
&=\int_{0}^1x^{k-1}\sum_{n=0}^\infty \frac{x^n}{n!}\,dx\\
&=\int_{0}^1x^{k-1}e^x\,dx\\
&\,\,\color{blue}{=(-1)^k(k-1)!\left(1-e\sum_{j=0}^{k-1}\frac{(-1)^j}{j!}\right)}
\end{align*}
The last line can be derived using integration by parts together with a proof based upon induction on $k$.
