# Formula for $\sum \limits_{n=0}^{\infty} \frac{1}{(n+a)!}$

Is there a closed form for the infinite sum $$\sum \limits_{n=0}^{\infty} \frac{1}{(n+a)!} \mathrm{?}$$ where a is an integer greater than or equal to $0$.

When $a=0$, the sum is just the series expansion for $e$, $\sum \limits_{n=0}^{\infty} \frac{1}{n!}=e$

When $a=1$, the sum can be rewritten as $\sum \limits_{n=0}^{\infty} \frac{1}{(n+1)!}= \sum \limits_{n=1}^{\infty} \frac{1}{n!} =\frac{1}{1!}+\frac{1}{2!}+\frac{1}{3!}+ \dots=e-1$.

When $a=2$, the sum can be rewritten as $\sum \limits_{n=0}^{\infty} \frac{1}{(n+2)!}= \sum \limits_{n=2}^{\infty} \frac{1}{n!} =\frac{1}{2!}+\frac{1}{3!}+\frac{1}{4!}+\dots=e-2$.

When $a=3$, the sum can be rewritten as $\sum \limits_{n=0}^{\infty} \frac{1}{(n+3)!}= \sum \limits_{n=3}^{\infty} \frac{1}{n!} =\frac{1}{3!}+\frac{1}{4!}+\dots=e-\frac{5}{2}$.

$\vdots$

Thus, we can generalize that $$\sum \limits_{n=0}^{\infty} \frac{1}{(n+a)!}= \sum \limits_{n=a}^{\infty} \frac{1}{n!}.$$

Can this infinite sum be written in a closed form in terms of $a$? Following the pattern the general form would have some value, dependent on $a$, subtracted from $e$.

• Can anyone explain why they voted to close this? – zz20s Feb 8 '16 at 13:03
• This is the value of the Mittag Leffler function $E_{1,a+1}$ at $z = 1$. Other than that, it is just what you wrote it is - $e$ minus a finite sum. – Hans Engler Feb 8 '16 at 13:46
• Right, my question is whether there exists a closed form for that finite sum. – zz20s Feb 8 '16 at 13:48

If you're familair with the gamma function:

$$\sum_{n=0}^{\infty}\frac{1}{(n+a)!}=\lim_{m\to\infty}\sum_{n=0}^{m}\frac{1}{(n+a)!}=$$ $$\lim_{m\to\infty}\frac{e\left(a!\Gamma(a+m+1,1)-a\Gamma(a,1)(a+m)!\right)}{a!(a+m)!}=$$ $$\frac{e}{a!}\lim_{m\to\infty}\frac{a!\Gamma(a+m+1,1)-a\Gamma(a,1)(a+m)!}{(a+m)!}=$$ $$\frac{e}{a!}\lim_{m\to\infty}\left(\frac{a!\Gamma(1+a+m,1)}{(a+m)!}-a\Gamma(a,1)\right)=$$ $$\frac{e}{a!}\left(\lim_{m\to\infty}\frac{a!\Gamma(1+a+m,1)}{(a+m)!}-a\Gamma(a,1)\right)=$$ $$\frac{e}{a!}\left(a!\lim_{m\to\infty}\frac{\Gamma(1+a+m,1)}{(a+m)!}-a\Gamma(a,1)\right)=$$ $$\frac{e}{a!}\left(a!\cdot1-a\Gamma(a,1)\right)=e-\frac{e\Gamma(a,1)}{\Gamma(a)}$$

Not clearly more convenient, but an alternative form can be built by first establishing that:

$$\int_0^1\frac{x^n}{n!}e^{-x}\,dx=1-\frac{1}{e}\sum_{k=0}^n\frac{1}{k!}$$

[This is best done by induction - if we call the integral on the left $I_n$, integrating by parts quickly gives us $I_n=I_{n-1}-\frac{1}{e\cdot n!}$.]

Noting that $\sum_{k=0}^{\infty}\frac{1}{k!}=e$, this can be recast as:

$$\int_0^1\frac{x^n}{n!}e^{-x}\,dx=\frac{1}{e}\sum_{k=n+1}^{\infty}\frac{1}{k!}$$

and shifting around the $n!$ and $e$ terms gives that

$$\sum_{n=a}^{\infty}\frac{1}{n!}=\int_0^1\frac{x^{a-1}}{(a-1)!}e^{1-x}\,dx$$