# Finite series with reciprocal factorials

I asked the question at MSE https://math.stackexchange.com/questions/982388/simple-finite-series-with-reciprocal-factorials but got no answer or comment (it is not a homework).

I'm trying to find the following sum: $$\sum_{k=0}^n{1\over(n-k)!}{x^{k+2}\over k+2}.$$ The most obvious way is to differentiate wrt to $x$ leading to $$\sum_{k=0}^n{1\over(n-k)!}{x^{k+1}}=\frac{1}{n!}e^{\frac{1}{x}} x^{n+1} \Gamma \left(n+1,\frac{1}{x}\right)$$ (according to Mathematica) but I don't see how to integrate the right side to get the original sum ($\Gamma$ is the incomplete Gamma function).

Idea I for a solution: Isn't it possible to insert an elementary auxiliary function $f(y)$ to the first equation, whose integral wrt to $y$ will cancel the ${1\over k+2}$ term, then find the sum, differentiate it wrt to $y$ and set the auxiliary function to one?

Such a function exists $f(y)=y^{-k/(k + 2)}$ but the sum is then unsummable.

Idea II for a solution: Multiply/divide the summands by both $n!$ and $k!$ to write it as a combinatorial series (omitting some constant functions of $n$ and $x$) $$\sum_{k=0}^n\binom{n+2}{k+2}{x^{k}}k!(k+1).$$ A lot is known about series with binomial coefficients (see here http://www.math.wvu.edu/~gould/) but I found no way out of it. I think the last form indicates that the sum can be written in terms of elementary functions (but it may be quite a huge expression for an arbitrary $k$).

According to Maple your sum is:

$$\,{\frac {{x}^{2}{_3F_1(1,2,-n;\,3;\,-x)}}{2n!}}$$

1/2/n!*x^2*hypergeom([1, 2, -n],[3],-x)

• Impressive, I gave up on Maple some time ago after some bad experience but... Thank you! – user155002 Oct 24 '14 at 14:46
• This hypergeometric expression says no more and no less than that the coefficient of $x^{k+2}$ in the sum is $$\frac{1}{(n-k)!\,(k+2)}.$$ – Ira Gessel Oct 24 '14 at 16:48
• Fair enough, but how is different from so-called elementary functions? I mean, the important thing except adding a name to a series $\sum_{k=0}^\infty{x^k\over k!}$ is that one can prove the properties of $e^x$ and relate it to other functions (as many as possible). Then it becomes useful and somehow satisfactory. So I don't see a difference between $e^x$ and $_3F_1$ from this point of view. – user155002 Oct 24 '14 at 20:32

You might also note that the generating function of the sequence of your sums is $$g(t) = \sum_{n=0}^\infty t^n \sum_{k=0}^n \dfrac{x^{k+2}}{(n-k)!(k+2)} = - e^t \left(\dfrac{x}{t} + \dfrac{\ln(1-tx)}{t^2}\right)$$

• That is remarkable. – user155002 Oct 24 '14 at 23:32