Find the following sum:
$$\sum_{k=1}^\infty \frac{1}{k^2}\frac{x^k}{k!}$$
where $x$ is a real number. This is a power series in $x$. In particular, I'm interested in the case $x>0$.
Disclaimer: This is not a homework exercise, I do not know if a closed form solution exists. If it doesn't, exist, then an approximation in terms of well-known functions (not the all-mighty general hypergeometric $_pF_q$, something simpler please) would be desired.
We know that
$$e^x=1+\sum_{k=1}^\infty\frac{x^k}{k!}$$
Divide both sides to get
$$\frac{e^x}x=\frac1x+\sum_{k=1}^\infty\frac{x^{k-1}}{k!}$$
Integrate both sides to get
$$\int_1^x\frac{e^y}ydy+c_1=\ln(x)+\sum_{k=1}^\infty\frac1k\frac{x^k}{k!}$$
Divide both sides again.
$$\frac1x\left(\int_1^x\frac{e^y}ydy+c_1-\ln(x)\right)=\sum_{k=1}^\infty\frac1k\frac{x^{k-1}}{k!}$$
Integrate both sides to get
$$\int_1^x\frac1z\left(\int_1^z\frac{e^y}ydy+c_1-\ln(z)\right)dz+c_2=\sum_{k=1}^\infty\frac1{k^2}\frac{x^k}{k!}$$
which is quite messy?
This is a trick i discovered recently that allows you to express this kind of sum as an integral (not sure how helpful you'll find this): We have
$$ \sum_{k=1}^{\infty} \frac{1}{k^2}\frac{x^k}{k!} = \sum_{k=1}^{\infty}\frac{x^k}{k!} \int_{0}^{1}\int_{0}^{1} y^{k-1}z^{k-1} \mathrm{d}y\;\mathrm{d}z$$ As $x > 0$, we may interchange summation and integration to obtain
$$ \int_{0}^{1}\int_{0}^{1}\frac{1}{yz}\sum_{k=1}^{\infty}\frac{(xyz)^k}{k!} \mathrm{d}y\;\mathrm{d}z = \int_{0}^{1}\int_{0}^{1}\frac{\exp(xyz)-1}{yz} \mathrm{d}y\;\mathrm{d}z$$
http://math.stackexchange.com/questions/1711318/show-sum-k-1-infty-frack2k-2-mathrme/1711321#1711321
– orion Nov 08 '16 at 22:40