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Playing around with the harmonic number $H(n) = 1+ \frac{1}{2} + \frac{1}{3} + ... + \frac{1}{n}$ and its asymptotic approximation

$$H(n\to\infty) \simeq \gamma + \log(n) +\frac{1}{2n} -\frac{1}{12 n^2} + ...$$

I started with first order comparisions studying these sums

$$s_H = \sum _{k=1}^{\infty } \frac{(-1)^k H_{k-1}}{k}$$ $$s_L= \sum _{k=1}^{\infty } \frac{(-1)^k (\log (k)+\gamma )}{k}$$

Notice the shift in the argument of $H(k)$.

Mathematica tells me that

$$s_H = \frac{1}{2}\log ^2(2)$$ $$s_L = - \frac{1}{2}\log ^2(2)$$

Normally we would expect that the sum using $H$ should be comparable to that using an approximation of $H$. But this expectation is not justified at all here.

Hence we are "naturally" led to ask for this sum

$$s_{HL} = \sum _{k=1}^{\infty } \frac{(-1)^k \left(H_{k-1}+\left(\log (k)+\gamma \right)\right)}{k}$$

Here Mathematica finds that

$$s_{HL} = 0\tag{*}$$

I wouldn't have guessed that from looking at the summands.

Hence an independent proof of $(*)$ would be desirable.

Remark: this problem is related to A closed form of the family of series $\sum _{k=1}^{\infty } \frac{\left(H_k\right){}^m-(\log (k)+\gamma )^m}{k}$ for $m\ge 1$

Dr. Wolfgang Hintze
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  • you are not rearranging the order of summation of the terms, you are just adding two convergent series. To see this you can use the definition of the values of the series as a limit of sums – Masacroso Aug 11 '19 at 09:54
  • @ masacroso you are right. Thank you. I have rephrased the question. – Dr. Wolfgang Hintze Aug 11 '19 at 11:04
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    Why wouldn't you just compute the two sums directly? Isn't this itself the obvious approach to tackling the series? – Simply Beautiful Art Aug 11 '19 at 13:21
  • @ Simply Beautiful Art: That is exactly what I did in the first place (see my OP). But I didn't expect that the sum using the harmonic number is exactly the negative of the sum which uses the approximation of $H_n$. Here some delicate cancellation takes place. – Dr. Wolfgang Hintze Aug 12 '19 at 08:30

1 Answers1

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\begin{align} s_H&=\sum_{k=1}^\infty\frac{(-1)^kH_{k-1}}{k}\\ &=\sum_{k=1}^\infty\frac{(-1)^kH_{k}}{k}-\sum_{k=1}^\infty\frac{(-1)^k}{k^2}\\ &=\frac12\ln^2(2)+\operatorname{Li}_2(-1)-\operatorname{Li}_2(-1)\\ &=\frac12\ln^22 \end{align} where we used the generating function $\ \sum_{k=1}^\infty\frac{x^k H_k}{k}=\frac12\ln^2(1-x)+\operatorname{Li}_2(x)$

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From this solution , we proved $\ \int_0^\infty \ln x\ e^{-kx}\ dx=-\frac{\ln k+\gamma}{k}$ , so we can write

\begin{align} s_L&=\sum_{k=1}^\infty\frac{(-1)^k(\ln k+\gamma)}{k}\\ &=-\int_0^\infty\ln x\sum_{k=1}^\infty(-e^{-x})^k\ dx\\ &=\int_0^\infty\frac{e^{-x}\ln x}{1+e^{-x}}\ dx \end{align}

which I think manageable to prove it $-\frac12\ln^22$

Ali Shadhar
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    For the last integral is enough to consider: $$I(s)=\int_0^\infty \frac{x^{s-1}}{e^x+1}dx=\eta(s)\Gamma(s)$$ $$\Rightarrow I'(s)=\int_0^\infty \frac{x^{s-1}\ln x}{e^x+1}dx=\frac{d}{ds}\left(\eta(s)\Gamma(s)\right)$$ And the result: $-\frac12 \ln^2 2$ follows by plugging $s=1$ and using https://en.wikipedia.org/wiki/Dirichlet_eta_function#Derivatives. – Zacky Aug 11 '19 at 17:22
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    and solution is complete :) – Ali Shadhar Aug 11 '19 at 17:26
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    But now that I think of it, we don't need the integral representation, since $\sum_{k=1}^\infty\frac{(-1)^k \ln k}{k}$ is actually $\eta'(1)$. Well both ways ends to the same result. – Zacky Aug 11 '19 at 17:30
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    That's a way to go too. – Ali Shadhar Aug 11 '19 at 17:34
  • @ Ali Shather Elegant and efficient, as always. May I draw your attention to my older question, on which you are herewith invited to test your capabilities? (https://math.stackexchange.com/questions/2582497/a-closed-form-of-the-family-of-series-sum-k-1-infty-frac-lefth-k-rig) – Dr. Wolfgang Hintze Aug 12 '19 at 08:52
  • @Dr. Wolfgang Hintze thank you for your kind words, it means a lot coming from an experienced person like you. That problem is really interesting I will try it and see what I get. – Ali Shadhar Aug 12 '19 at 19:55