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I stumbled on this problem in the wake of the discussion https://math.stackexchange.com/a/3553902/198592

Can you make sense of the equation in the question involving the harmonic number with a rational index $H_{\frac{2}{3} k}$ although the series is not convergent?

Hint: find the generating function of $g(z) = \sum_{k=1}^{\infty}z^k H_{\frac{2 k}{3}}$ and interpret the sum as the limit $z\to -1$.

Dr. Wolfgang Hintze
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  • Nice problem +1. Have you tried using the integral representation of the harmonic number ? – Ali Shadhar Feb 20 '20 at 21:49
  • Some users say that this sum does not converge and I also asked in a post yesterday about Grandi series if it converges to 1/2 or diverges and got different solutions and opinions. If grandi series diverges then this sum diverges due to the relation between the two sums. Grandi series seems controversial. – Ali Shadhar Feb 22 '20 at 22:43
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    @ Ali Shather It was, of course, my intention. I wanted to give an example of an extension of the common type of question to divergent sums. There should have been no misunderstanding as I have, in the OP, pointed out clearly the divergence of the sum and asked to "make sese" of the sum. But maybe I should better have put the word "prove" in hyphens ... – Dr. Wolfgang Hintze Feb 23 '20 at 11:40

1 Answers1

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Start with using the integral representation of the harmonic number $H_n=\int_0^1\frac{1-x^n}{1-x}\ dx$ we have

$$\sum_{k=0}^\infty(-1)^k H_{\frac{2k}{3}}=\int_0^1\frac{1}{1-x}\sum_{k=0}^\infty((-1)^k-(-x^{\frac23})^n)\ dx$$

$$\int_0^1\frac{1}{1-x}\left(\frac12-\frac{1}{1+x^{\frac23}}\right)\ dx\overset{x\to x^3}{=}-\frac32\int_0^1\left(\frac{x}{1+x^2}-\frac{1}{1+x^2}+\frac{1}{1+x+x^2}\right)\ dx$$

$$=-\frac32\left[\frac12\ln(1+x^2)-\tan^{-1}x+\frac{2}{\sqrt{3}}\tan^{-1}\left(\frac{1+2x}{\sqrt{3}}\right)\right]_0^1$$

$$= -\frac{3}{4} \ln2+\frac{3 \pi }{8}-\frac{\pi }{2 \sqrt{3}}$$

Note that I used Grandi series $\sum_{k=0}^\infty (-1)^k=\frac12$.

Ali Shadhar
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  • @ Ali Shather +1 Nice derivation. I had obtained the g.f. using mathematica, and the limit gave the result. Only afterwards I discovered that the inital sum is not convergent but that the even and odd partial sums both go to $\infty$. But the mean of many partial sums approached the value given. This is the famous mechanism which you called Grandi series. The same would work, of course, for the simpler case of a divergent sum $\sum_{k=1}^\infty (-1)^k H_k =\frac{\log (1-z)}{z-1}|_{z\to-1}= -\frac{1}{2}\log(2)$ – Dr. Wolfgang Hintze Feb 21 '20 at 08:32
  • Thank you Wolf and nice example. By the way mathematica gave the sum divergent then it showed the numeric answer $-0.248663$. I saw such cases couple times. – Ali Shadhar Feb 21 '20 at 08:36
  • I suspect that Mathematica tries to find the g.f. and then applies the limit. That would be a reasonable procedere. At least there is an option to force a certain regularization scheme (e.g. Sum[(-1)^n , {n, 1, [Infinity]}, Regularization -> "Dirichlet"] $-\frac{1}{2}$). – Dr. Wolfgang Hintze Feb 21 '20 at 14:16
  • @Dr. Wolfgang Hintze I agree with you as it finally gives a result. – Ali Shadhar Feb 21 '20 at 16:32