Sum of series in the form $\sum^{\infty}_{n=0}\frac{1}{(n+k)\cdot n!}$ for some k>0

Question

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?

Answer 1

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.

Answer 2

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 $k$.