3
$\begingroup$

In the description of this question, it was established that \begin{align} \sum_{n=2}^{\infty} (\zeta(n)^{2}-1) &= \frac{7}{4} - \zeta(2) + 2 \sum_{m=1}^{\infty} \frac{H_{m-1- \frac{1}{m}} - H_{- \frac{1}{m}} - H_{m-1} }{m} \qquad(1) \end{align}

In the answer to that very same question, Fedor Petrov showed that an integral representation can be found for this sum. If we proceed from his derivation and use this calculation, we obtain \begin{align} \sum_{n=2}^{\infty}(\zeta(n)^{2}-1) &= 1 + \int_0^1 \sum_{k=0}^\infty\frac{x^{k}}{1+x+\ldots+x^{k+1}}dx \\ &= 1 + \sum_{k=0}^{\infty} \int_{0}^{1}\frac{x^{k}}{1+x+\ldots+x^{k+1}}dx \\ &= 1 - \sum_{k=0}^{\infty} \frac{H_{- \frac{1}{k+2}}}{k+2} \\ &= 1- \sum_{m=2}^{\infty} \frac{H_{-\frac{1}{m}}}{m}.\qquad (2) \end{align}

Now, we can equate the two expressions. Note that not all terms in the first sum can be separated, as only $\sum \frac{H_{-\frac{1}{m}}}{m}$ converges as a standalone series. Grouping like terms together, we find $$\sum_{m=2}^{\infty} \frac{H_{-\frac{1}{m}}}{m} = \frac{3}{4} - \zeta(2) + 2 \sum_{m=2}^{\infty} \frac{H_{m - 1- \frac{1}{m}} - H_{m-1} }{m}. \qquad \qquad (*)$$

What I find interesting here, is that (series involving) Harmonic numbers with both negative and positive fractional arguments can be related to one another.

Questions:

  1. Do identities like the $(*)$-marked equation appear in the literature?
  2. Can it be shown that one side of the equation amounts to the other side, only by means of algebraic manipulations and without invoking the aforementioned integral representation?
  3. I seem to have made some calculation error, because the sums don't appear to add up to the same number. Can this error be identified? Answer: this has been answered by Carlo Beenakker. The correct identity has now been established. The first two questions remain open.
$\endgroup$
2
+25
$\begingroup$

Q3: The first identity (1) is not correct, it should read \begin{align} \sum_{n=2}^{\infty}(\zeta(n)^{2} -1) &= \frac{7}{4} - \zeta(2) + 2\sum_{m=2}^{\infty} \frac{H_{m-1-\frac{1}{m}} - H_{-\frac{1}{m}} - H_{m-1}}{m} . \end{align} The final identity (*) then becomes $$ \sum_{m=2}^{\infty} \frac{H_{-\frac{1}{m}}}{m} = \frac{3}{4} - \zeta(2) + 2 \sum_{m=2}^{\infty} \frac{H_{m -1- \frac{1}{m}} - H_{m-1} }{m}. $$ Left-hand-side and right-hand-side both evaluate to $-1.4653$.

| cite | improve this answer | |
$\endgroup$
  • $\begingroup$ Ah I see, thank you. I appear to have made an error when applying the identity $H_{m} = \frac{1}{m} + H_{m-1}$ twice, though I'm still not entirely sure what went wrong. In any case, the first two questions remain now. $\endgroup$ – Max Muller Oct 9 at 12:02
  • $\begingroup$ the error in the application of that identity is that it does not help if applied to $H_{m-1-1/m}$, because of the $1/m$ term $\endgroup$ – Carlo Beenakker Oct 9 at 12:04
  • $\begingroup$ Alright, I should be more careful. If you have any ideas regarding questions (1) and (2), I would gladly learn about them. $\endgroup$ – Max Muller Oct 9 at 12:06

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.