Let f(x) = sum from n = 1 to infinity of x^n / n^2, the dilogarithm Li2(x). Is f differentiable at x = -1?

Step-by-step solution

1. 1

   Locate x = −1 relative to the interval of convergence. The ratio test gives radius $R=1$, and at both endpoints $\sum \tfrac{(\pm 1)^n}{n^2}$ converges absolutely (it is dominated by $\sum 1/n^2$). So $f$ is defined on the closed interval $[-1,1]$, and $x=-1$ is an endpoint — which is exactly why the question is not automatic. Inside $(-1,1)$ differentiability is free; at an endpoint it must be argued.

2. 2

   Differentiate term by term inside the interval. For $|x|<1$,

   $f'(x)=\sum_{n=1}^{\infty}\frac{n x^{n-1}}{n^2}=\sum_{n=1}^{\infty}\frac{x^{n-1}}{n}=-\frac{\ln(1-x)}{x}$

   using $\sum_{n\ge 1}\tfrac{t^n}{n}=-\ln(1-t)$ with $t=x$ and then dividing by $x$.

3. 3

   Test the differentiated series at the endpoint. Putting $x=-1$ into the derivative series gives

   $\sum_{n=1}^{\infty}\frac{(-1)^{n-1}}{n}=1-\frac12+\frac13-\frac14+\cdots$

   The alternating harmonic series converges (by the alternating series test) to $\ln 2$. This is the decisive difference from the case $x=+1$, where the same series becomes the divergent harmonic series and the derivative fails to exist.

4. 4

   Upgrade convergence to differentiability with Abel's theorem. Since the differentiated series converges at $x=-1$, it converges uniformly on $[-1,0]$, and a uniformly convergent series of derivatives may be integrated/differentiated term by term up to and including the endpoint. Hence $f$ is differentiable at $x=-1$ (one-sidedly from within $[-1,1]$, and two-sidedly if one uses the analytic continuation of $\operatorname{Li}_2$ past $-1$) with

   $f'(-1)=\sum_{n=1}^{\infty}\frac{(-1)^{n-1}}{n}=\ln 2$

5. 5

   Confirm with the closed form. The formula $f'(x)=-\tfrac{\ln(1-x)}{x}$ is continuous at $x=-1$ and gives

   $f'(-1)=-\frac{\ln 2}{-1}=\ln 2\approx 0.6931$

   agreeing with the series value. (For reference, $f(-1)=\sum \tfrac{(-1)^n}{n^2}=-\tfrac{\pi^2}{12}$.)

   $\boxed{\text{Yes; }f'(-1)=\ln 2\approx 0.693}$

6. 6

   Contrast the two endpoints. At $x=1$ the difference quotient $\tfrac{f(1)-f(x)}{1-x}=\sum_{n\ge 1}\tfrac{1+x+\cdots+x^{n-1}}{n^2}$ tends to $\sum \tfrac{1}{n}=\infty$, so $f$ has a vertical tangent there and is not differentiable. The single extra factor of $(-1)^n$ at $x=-1$ turns a divergent series into a convergent one — the whole answer turns on that.