Find the sum of x^n / n^2 for n from 1 to infinity.

Step-by-step solution

1. 1

   Find the radius of convergence first. The ratio test gives

   $\left|\frac{a_{n+1}}{a_{n}}\right|=\left|x\right|\frac{n^{2}}{(n+1)^{2}}\to\left|x\right|$

   so the series converges for $\left|x\right|<1$. At the endpoints it also converges: at $x=1$ it is the $p$-series with $p=2$, and at $x=-1$ it converges absolutely. The domain of the sum is therefore the closed interval $\left|x\right|\le 1$.

2. 2

   Build it from the geometric series by two integrations. Starting from $\sum_{n\ge 1}x^{n-1}=\dfrac{1}{1-x}$ and integrating once gives

   $\sum_{n=1}^{\infty}\frac{x^{n}}{n}=-\ln(1-x)$

   Dividing by $x$ and integrating a second time introduces the extra $n$ in the denominator:

   $\sum_{n=1}^{\infty}\frac{x^{n}}{n^{2}}=-\int_{0}^{x}\frac{\ln(1-t)}{t}\,dt$

3. 3

   Recognise that this integral has no elementary closed form. Unlike $\sum x^{n}/n$, the second antiderivative is not expressible with elementary functions. Mathematicians therefore give it a name — the dilogarithm:

   $\operatorname{Li}_{2}(x)=\sum_{n=1}^{\infty}\frac{x^{n}}{n^{2}}=-\int_{0}^{x}\frac{\ln(1-t)}{t}\,dt$

   Naming it is the honest answer, exactly as with the error function for $\int e^{-x^{2}}dx$.

4. 4

   Read off the two special values. At $x=1$ the series is the Basel sum:

   $\operatorname{Li}_{2}(1)=\frac{\pi^{2}}{6}$

   and at $x=-1$ the alternating version halves-and-flips it:

   $\operatorname{Li}_{2}(-1)=-\frac{\pi^{2}}{12}$

5. 5

   Check a third known value numerically. At $x=\tfrac12$ there is a classical identity $\operatorname{Li}_{2}\!\left(\tfrac12\right)=\dfrac{\pi^{2}}{12}-\dfrac{\ln^{2}2}{2}$. Summing 200 terms of the series gives $0.58224052646$, and the closed form gives $0.58224052647$ ✓ — confirming both the series and the naming convention.