Is it possible to work the following summation using standard methods?

Question

Is it possible to get a summation without the use of the Lerch transcendent?

$$\sum_{k=1}^x \frac{a^{2k}}{2k}$$

Summation Result using Mathematica

Answer 1

Disclaimer: There is no closed form I know of.

Recall the geometric series:

$$\frac{1-r^n}{1-r}=\sum_{k=0}^{n-1}r^k$$

Integrate both sides to get

$$\int_0^{a^2}\frac{1-r^n}{1-r}\ dr=\sum_{k=0}^{n-1}\int_0^{a^2}r^k\ dr=\sum_{k=0}^{n-1}\frac{(a^2)^{k+1}}{k+1}=\sum_{k=1}^n\frac{a^{2k}}k$$

Divide both sides by two and you get

$$\sum_{k=1}^n\frac{a^{2k}}{2k}=\frac12\int_0^{a^2}\frac{1-r^n}{1-r}\ dr$$