# Sum of a power series with a parameter

Does anybody know the sum

$$\sum_{n=0}^{\infty}\frac{x^{n}}{(n+a)n!}=f(x,a)$$

here $a$ is a number $a>0$. A hint please ? :D

If $a=1$ I believe $f(x,1)=(e^{x}-1)/x$

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Hint: Study the function $x\mapsto x^af(x,a)$. I'm afraid, however, that the result is a special function... – Tom-Tom Feb 7 '14 at 13:56
It can be expressed in terms of Confluent hypergeometric function of the First Kind. $$f(x,a) = \frac{1}{a} {}_1\!F_1( a; a+1; x)$$ – achille hui Feb 7 '14 at 19:36

For $a\gt0$, \begin{align} af(x,a) &=a\sum_{n=0}^\infty\frac{x^n}{(n+a)n!}\\ &=1+\sum_{n=1}^\infty\left(\frac1n-\frac1{n+a}\right)\frac{x^n}{(n-1)!}\\ &=e^x-x\sum_{n=0}^\infty\frac{x^n}{(n+a+1)n!}\\ &=e^x-xf(x,a+1)\tag{1} \end{align} Reversing $(1)$ yields $$f(x,a)=\frac{e^x-(a-1)f(x,a-1)}{x}\tag{2}$$ Evaluating $(1)$ directly gives \begin{align} f(x,1) &=\sum_{n=0}^\infty\frac{x^n}{(n+1)n!}\\ &=\frac1x\sum_{n=0}^\infty\frac{x^{n+1}}{(n+1)!}\\ &=\frac{e^x-1}{x}\tag{3} \end{align} $f(x,a)$ can be computed for higher $a\in\mathbb{Z}$ using recursion and $(2)$.
I think you should rewrite your function in a better way... $$f(x,a) = \sum \frac{x^n}{(n+a)n!} = \sum \frac{x^n}{n \cdot n!} + \sum \frac{x^n}{a \cdot n!} = S + \frac{1}{a}\sum \frac{x^n}{n!} = S + \frac{e^x}{a}.$$ If you differentate $S$, you'll find that $S' = \sum \frac{n \cdot x^{n-1}}{n \cdot n!} = \sum \frac{x^{n-1}}{n!}$, so $xS' = \sum \frac{x^n}{n!} = e^x$. Now ... since $xS' = e^x, S = \int e^x / x = Ei(x)$. So $f(x,a) = \sum \frac{x^n}{(n+a)n!} = e^x/a + Ei(x)$.
$\frac1{n+a}\ne\frac1n+\frac1a$ – robjohn Feb 7 '14 at 20:59