Compute $\sum_{k=0}^\infty \frac{t^{k+1}}{(k+1)^2}$ for $t\in (0,1)$.

Question

How can I Compute $$\sum_{k=0}^\infty \frac{t^{k+1}}{(k+1)^2}$$ for $t\in (0,1)$ ? I tried many things, but not conclusive.

Answer 1

Let $I(t)$ you sum. Then, $$\frac{\mathrm d I(t)}{\mathrm d t}=\sum_{k=0}^\infty \frac{t^k}{k+1}$$ and thus $$t\frac{\mathrm d I(t)}{\mathrm d t}=\sum_{k=0}^\infty \frac{t^{k+1}}{k+1}\implies \frac{\mathrm d }{\mathrm d t}\left(t\frac{\mathrm d I(t)}{\mathrm d t}\right)=\sum_{k=0}^\infty t^k=\frac{1}{1-t}.$$ Therefore, you get the following differential equation :

$$t\ddot I(t)+\dot I(t)=\frac{1}{1-t},$$ which is not complicate to solve.

Answer 2

One may note that we have the polylogarithm:

$$\operatorname{Li}_s(z)=\sum_{n=0}^\infty\frac{z^{n+1}}{(n+1)^s}$$

And it comes with known recursive formula $\operatorname{Li}_{s+1}(z) = \int_0^z \frac {\operatorname{Li}_s(t)}{t}\,\mathrm{d}t$ and $\operatorname{Li}_0(z)=\frac z{1-z}$.

Answer 3

We have,

$$\int_{0}^{t} x^k dx=\frac{t^{k+1}}{k+1}$$

So our series is,

$$\int_{0}^{t} \sum_{k=0}^{\infty} \frac{x^k}{k+1} dx $$

$$=\int_{0}^{t} \frac{1}{x} \sum_{k=1}^{\infty} \frac{x^k}{k} dx$$

$$=\int_{0}^{t} \frac{1}{x} \left(\int_{0}^{x} \sum_{k=0}^{\infty} w^k dw \right) dx$$

Now recognize the geometric series to get,

$$=\int_{0}^{t} -\frac{\ln (1-x)}{x} dx$$

This is $\text{Li}_2(t)$.

Answer 4

By the change of index $n=k+1$ one gets $$ \sum_{k=0}^\infty \frac{t^{k+1}}{(k+1)^2}=\sum_{n=1}^\infty \frac{t^n}{n^2},\qquad |t|<1, \tag1 $$ then recognizing the power series of the dilogarithm function (see here): $$ \text{Li}_2(t):=\sum_{n=1}^\infty \frac{t^n}{n^2}=-\int_0^t\frac{\ln(1-x)}x\:dx,\qquad |t|<1, \tag1 $$ which admits many interesting properties, for example with $|z|<1$, $$\begin{align} &\operatorname{Li}_2(z)+\operatorname{Li}_2(-z)=\frac{1}{2}\operatorname{Li}_2(z^2)\\ &\operatorname{Li}_2(1-z)+\operatorname{Li}_2\left(1-\frac{1}{z}\right)=-\frac{\ln^2z}{2} \\ &\operatorname{Li}_2(z)+\operatorname{Li}_2(1-z)=\frac{{\pi}^2}{6}-\ln z \cdot\ln(1-z) \\ &\operatorname{Li}_2(-z)-\operatorname{Li}_2(1-z)+\frac{1}{2}\operatorname{Li}_2(1-z^2)=-\frac {{\pi}^2}{12}-\ln z \cdot \ln(1+z). \end{align} $$