Find a function for the power series $\sum_{n=0}^\infty \frac{z^{3n+1}}{3n+1}$ I have to find the function for $\sum_{n=0}^\infty \frac{z^{3n+1}}{3n+1}$
I found on Finding a function corresponding to the complex power series that I could uset the fact that  $\sum_{n=0}^\infty z^n = \frac{1}{1-z} \ \forall |z| < 1$. But if I make the substitution $m = 3n + 1$,  I get $\sum_{m=1}^\infty \frac{z^m}{m}$. Because now the sum goes from $1$ to $\infty$ and not from $0$, as far as I know, I cannot use the formula for the geometric series. Am I wrong about my assumption or do I have to determine it some other way?
 A: Denote
$$
f(z) = \sum_{n=0}^\infty \frac{z^{3n+1}}{3n+1}.
$$
Now you can differentiate the series and obtain
$$
f'(z) = \sum_{n=0}^\infty z^{3n} = \frac{1}{1 - z^3}.
$$
After this you can find $f(z_0)$ at any point $z_0$ by integrating and using the fact that $f(0) = 0$
$$
f(z_0) = \int_{0}^{z_0} f'(z) \, dz.
$$
A: $\displaystyle\sum_\limits{n=0}^\infty\dfrac{z^{3n+1}}{3n+1}=\sum_\limits{n=0}^\infty\int_0^z\!\!x^{3n}\,\mathrm dx=\int_0^z\!\sum_\limits{n=0}^\infty x^{3n}\,\mathrm dx=\int_0^z\!\!\!\dfrac1{1-x^3}\,\mathrm dx$
Can you calculate the last definite integral ?
Note that
$\dfrac1{1-x^3}=\dfrac13\!\cdot\!\dfrac1{1-x}+\dfrac16\!\cdot\!\dfrac{1+2x}{1+x+x^2}+\dfrac12\!\cdot\!\dfrac1{\left(x+\frac12\right)^2+\frac34}\;.$
Moreover ,
$\displaystyle\int_0^z\dfrac1{1-x}\,\mathrm dx=\big[-\ln(1-x)\big]_0^z=-\ln(1-z)\;\;,$
$\displaystyle\int_0^z\dfrac{1+2x}{1+x+x^2}\,\mathrm dx=\big[\ln\left(1+x+x^2\right)\big]_0^z=\ln\left(1+z+z^2\right)\;\;,$
$\displaystyle\int_0^z\dfrac1{\left(x+\frac12\right)^2+\frac34}\,\mathrm dx=\left[\dfrac2{\sqrt3}\arctan\left(\dfrac{2x+1}{\sqrt3}\right)\right]_0^z=$
$\qquad=\dfrac2{\sqrt3}\arctan\left(\dfrac{2z+1}{\sqrt3}\right)-\dfrac{\pi}{3\sqrt3}\;\;.$
