Proving that $\lim_{x\rightarrow0}\frac{H_x}x=\frac{\pi^2}6$

Question

Yes, you read that correctly. It can be proven that $$\lim_{x\rightarrow0}\frac{H_x}x=\frac{\pi^2}6$$Where $H_x$ is the $x$th harmonic number and is analytically continued to all positive reals. To prove this, we use a formula in the OP of this question where it is stated that $$H_x=\sum_{k=1}^\infty\frac{x}{k(x+k)}$$And the rest is obvious. But this proof is very boring and not intuitive. So I want to see other methods of proving this limit.

Answer 1

\begin{align*} \mathop {\lim }\limits_{x \to 0} \frac{{H_x }}{x} &= \mathop {\lim }\limits_{x \to 0} \frac{1}{x}\int_0^1 {\frac{{1 - t^x }}{{1 - t}}{\rm d}t} = \int_0^1 {\mathop {\lim }\limits_{x \to 0} \frac{{1 - t^x }}{x}\frac{1}{{1 - t}}{\rm d}t} = - \int_0^1 {\frac{{\log t}}{{1 - t}}{\rm d}t} \\ & = - \int_0^1 {\frac{{\log (1 - t)}}{t}{\rm d}t} = \sum\limits_{n = 1}^\infty {\frac{1}{n}\int_0^1 {t^{n - 1} {\rm d}t} } = \sum\limits_{n = 1}^\infty {\frac{1}{{n^2 }}} = \frac{{\pi ^2 }}{6} \end{align*}

Answer 2

$$H_x=\sum_{k=1}^\infty\frac{x}{k(x+k)}=\psi ^{(0)}(x+1)+\gamma$$ Using series $$\psi ^{(0)}(x+1)=-\gamma +\frac{\pi ^2}{6}x+\frac{ \psi ^{(2)}(1)}{2}x^2+\frac{\pi^4 }{90}x^3+O\left(x^4\right)$$ $$\frac{H_x}x=\frac{\pi ^2}{6}+\frac{ \psi ^{(2)}(1)}{2}x+\frac{\pi ^4 }{90}x^2+O\left(x^3\right)$$ is a good approximation for $0\leq x \leq \frac 16$.

A much better approximation is $$\frac{H_x}x=\frac{\pi ^2}{6}-\frac{45 x \psi ^{(2)}(1)^2}{2 \pi ^4 x-90 \psi ^{(2)}(1)}$$ whose error is $\frac {x^3}{16}$.

Answer 3

Using the hypergeometric representation of the Harmonic numbers: $$ \lim_{x\to 0}\frac{H_x}{x}=\lim_{x\to 0}\frac{x{_3F_2}\left({1,1,1-x\atop 2,2};1\right)}{x}={_3F_2}\left({1,1,1\atop 2,2};1\right). $$ Then using the generalized hypergeometric series we find $$ {_3F_2}\left({1,1,1\atop 2,2};1\right)=\sum_{n=0}^\infty \frac{n!\,n!\,n!}{(n+1)!\,(n+1)!\, n!}=\sum_{k=0}^\infty\frac{1}{(n+1)^2}=\frac{\pi^2}{6}. $$

Edit:

The OP is interested in plotting this function so here is a plot of the hypergeometric representation for $H_x$ on $x\in(-1,5)$: