6
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I have not the ability to compute more than four digits of

$$\sum_{n=1}^\infty \frac{1}{n^2 H_n^{(\ln n)}}$$

$H_n^{(m)} = \sum_{k=1}^n \frac{1}{k^m}$ is the generalized harmonic number.

I know this is the weirdest sum and it offers me no actual interest. All I know is that the decimal number starts off as $1.414...$ and I want to settle my mind that it is not actually $\sqrt 2$. That would be crazy. I have no reason to expect it. I just want some confirmation.

My calculations were from Desmos here https://www.desmos.com/calculator/helb1dgf1g.

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  • $\begingroup$ $\sum_{k=1}^n{n^{-\ln(k)}}=\sum_{k=1}^n{k^{-\ln(n)}}$. Is a pretty weird property. Not sure if it's helpful but this $H_n^{(\ln n)}$ has ... weird properties. $\endgroup$ – Mason Jul 3 '18 at 23:30
  • $\begingroup$ Out of curiosity, have you considered using any extrapolation techniques to estimate a sum for the series? Compute for e.g. $n=100, n=316, n=1000, n=3162, n=10000$ or for $n=1000, n=2000, n=3000, \ldots$ or something similar, and try and fit appropriately. Since $H_n^{(\ln n)}\in O(1)$ one would expect the partial sums to satisfy $S_n=S-\frac{C}{n}+o(n^{-1})$, where $S$ is the value of the infinite sum, and you could extrapolate using this to get a (probably) better estimate for $S$. $\endgroup$ – Steven Stadnicki Jul 4 '18 at 0:02
  • $\begingroup$ One thing is clear, if this thing does converge to $\sqrt{2}$, it does so VERY slowly. At 10000 terms I get 1.41404032445, at 20000 terms I get 1.41409025035, and $\sqrt{2}$ is 1.41421356237. $\endgroup$ – AlexanderJ93 Jul 4 '18 at 0:04
  • $\begingroup$ Just throwing an idea at the wall here... is it possible for us to square it? math.stackexchange.com/questions/813684/square-of-a-series $\endgroup$ – Mason Jul 4 '18 at 0:10
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    $\begingroup$ Note that the comments by AlexanderJ93 and @OscarLanzi above together imply that $\sqrt2$ won't be reached, as the sum is more than $1/20000$ from $\sqrt2$ after the first $20000$ terms. $\endgroup$ – joriki Jul 4 '18 at 2:18

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