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.
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$\begingroup$ Comments have been moved to chat; please do not continue the discussion here. Before posting a comment below this one, please review the purposes of comments. Comments that do not request clarification or suggest improvements usually belong as an answer, on Mathematics Meta, or in Mathematics Chat. Comments continuing discussion may be removed. $\endgroup$– Pedro ♦Commented Mar 14, 2023 at 12:14
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3 Answers
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\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*}
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$\begingroup$ Much more beautiful then my proof! I tried to do it that way but I didn't know that $\lim_{x\rightarrow0}\frac{1-t^x}x=-\ln t$. $\endgroup$ Commented Mar 9, 2023 at 17:40
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2$\begingroup$ @KamalSaleh Note that $$ \mathop {\lim }\limits_{x \to 0} \frac{{1 - t^x }}{x} = - \mathop {\lim }\limits_{x \to 0} \frac{{t^x - 1}}{x} = - \mathop {\lim }\limits_{x \to 0} \frac{{{\rm e}^{x\ln t} - 1}}{x} = - \ln t \cdot \mathop {\lim }\limits_{x \to 0} \frac{{{\rm e}^{x\ln t} - 1}}{{x\ln t}} = - \ln t. $$ $\endgroup$– GaryCommented Mar 9, 2023 at 21:28
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$$H_x=\sum_{k=1}^\infty\frac{x}{k(x+k)}=\psi ^{(0)}(x+1)+\gamma$$ 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 $\frac {x^3}{16}$.
answered Mar 9, 2023 at 5:22
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$\begingroup$ Taylor series expansion almost always works for some reason. But how did you get those coefficients? I am assuming that the coefficients for odd powers $n$ are $\zeta(2n-1)$ $\endgroup$ Commented Mar 9, 2023 at 5:27
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$\begingroup$ @KamalSaleh. Yes and yes. The coefficients are obtained by successive differentiations of the digamma function. Have a look at reference.wolfram.com/language/ref/PolyGamma.html $\endgroup$ Commented Mar 9, 2023 at 5:38
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$\begingroup$ Okay, thanks! I just realized that this is the expansion of the di-gamma function and not the Harmonic function. Makes much more sense to me now. $\endgroup$ Commented Mar 9, 2023 at 5:51
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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 $x\in(-1,5)$:
answered Mar 9, 2023 at 23:35
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1$\begingroup$ Just realized that $(a)_n$ is the pochhammer symbol. Thanks for your answer! +1 $\endgroup$ Commented Mar 10, 2023 at 2:21
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$\begingroup$ @KamalSaleh Yes, where $(1)_n=n!$ and $(2)_n=(n+1)!$ $\endgroup$ Commented Mar 10, 2023 at 2:25
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$\begingroup$ So what is $(1-x)_n$? I am struggling to graph this on Desmos. $\endgroup$ Commented Mar 10, 2023 at 2:30
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$\begingroup$ @KamalSaleh Demos may not be the best software to compute this function. Nonetheless, $(1-x)_n=\Gamma(1-x+n)/\Gamma(1-x)$. $\endgroup$ Commented Mar 10, 2023 at 2:34
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1$\begingroup$ @KamalSaleh At the end of the day, all the answers to this question can all be transformed into each other. They are all just slightly different perspectives of the same problem. In the case of the hypergeometric series in my post, it can be shown that this series boils down to the series in your original post with the same convergence properties. $\endgroup$ Commented Mar 12, 2023 at 20:10