Proving that $H'_0=\frac{\pi^2}6$

Question

I was taking the derivative of $$H_x=\sum_{k=1}^\infty\frac x{k(k+x)}$$ when I found out that $$H_0'=\frac{\pi^2}6$$Simply taking the derivative and plugging $x=0$ is not intuitive to me. What are other ways to prove that $H'_0=\frac{\pi^2}6$?

Answer 1

If you do not like to take the term-by-term derivative of the series, you can come back to the definition of derivative $H'_0 = \lim_{x\to 0} (H_x - H_0)/x$. Since $$ S_x = \sum_{k=1}^\infty \frac{1}{k\,(k+x)} $$ is an absolutely convergent sum for $x>-1$ and $H_x = x\,S_x$, one deduces that $H_0 = 0$. Hence for any $x\neq 0$, $$ \frac{H_x - H_0}{x} = \frac{x\,S_x - 0}{x} = S_x $$ and it follows easily by taking the limit in this sum that $H'_0 = S_0 = \sum_{k=1}^\infty \frac{1}{k^2}$ which is well-known to be equal to $\pi^2/6$.

Answer 2

Here's another approach. $$\lim_{x\rightarrow0}\frac{H_x}x=\frac{\pi^2}6$$By L'Hopital's rule $$\lim_{x\rightarrow0}H'(x)=H'(0)=\frac{\pi^2}6$$