Infinite Sum with differential operator How would one suggest to calculate the following sum? 
$\sum^{∞}_{n=1}\partial_{x}^{2n}(\frac{\pi}{2x}Erf[\frac{cx}{2}])=?$
where c is just a constant.
cheers.
 A: First,
I don't see why you need partial derivatives,
since you are only differentiating
w.r.t. x.
Since
$$\begin{align}
\operatorname{Erf}(z) & = \frac{2}{\sqrt{\pi}}\int_0^z e^{-t^2}dt, \\
\operatorname{Erf}'(z) & = \frac{2}{\sqrt{\pi}} e^{-z^2}.
\end{align}$$
You can then show by induction
that
$\operatorname{Erf}^{(n)}(z)
= \frac{2}{\sqrt{\pi}} e^{-z^2}P_n(z)
$
where $P_n(z)$ is a polynomial of degree $n-1$ in $z$.
Then,
apply Leibniz's rule
$$(f(z)g(z))^{(n)}
=\sum_{k=0}^n \binom{n}{k} f^{(k)}(z)g^{(n-k)}(z)
$$
after getting the formula for
$\left(\frac1{z}\right)^{(n)}$.
A: The series doesn't seem to converge anywhere.
First, let's look at whether it converges at $x=0$. The power series for $Erf$ is
$$
Erf(z)=\frac{2}{\sqrt{\pi}}\sum_{k=0}^\infty \frac{(-1)^k}{(2k+1)k!}z^{2k+1}
$$ from which
$$
\frac{\pi}{2x}Erf\left[\frac{cx}{2}\right]=\frac{c\sqrt{\pi}}{2}\sum_{k=0}^\infty \frac{(-1)^k}{(2k+1)k!}\left(\frac{cx}{2}\right)^{2k}
$$
Now to work out the $n$th term of your sum, evaluated at $x=0$, just the $k=n$ term of the series contributes, to get
$$
\left.\left(\frac{d}{dx}\right)^{2n}\left(\frac{\pi}{2x}Erf\left[\frac{cx}{2}\right]\right)\right|_{x=0}=\frac{c\sqrt{\pi}}{2}\frac{(-1)^n (2n)!}{(2n+1)n!}\left(\frac{c}{2}\right)^{2n}
$$
The terms on the right hand side here blow up for any $c$, so the sum doesn't converge at $x=0$.
It seems to be just as bad, or worse when $x$ is very large: $Erf$ is exponentially close to unity at large arguments, so the $n$th term of the series is roughly $\frac{\pi}{2}\frac{(2n)!}{x^{2n+1}}$ for sufficiently large $x$.
So I would bet (though what's here is not a complete proof except at $x=0$) that the series converges nowhere, excepting for the trivial $c=0$ case. On the other hand, if you modify the series with an extra $n!$ in the denominator, it becomes much more interesting! For example, at $x=0$, the modified sum converges for $|c|\leq 1$, and evaluates to $\frac{\sqrt{\pi}}{2}\sinh^{-1}(c)$.
