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Numerically, it would seem the following identity holds true:

$$\frac{6}{7}=\lim_{n\to\infty}\sqrt[n]{\sum_{k=3}^\infty{\left(k-\sum_{j=1}^{k}\frac{1}{j}\right)^{-n}}}$$

Down below I have proven that the right side of the equation exists between 0 and 1.

Let $H_k$ be the $k$th harmonic number. Consider the sequence $x_k=k-H_k$. It is easy to show that $x_k$ is increasing. Therefore, $x_k>1$ for all $k \geq 3$ since $x_3 = \frac{7}{6} > 1$. It follows that $x_k^{-n}$ decreases as $n$ increases if $k \geq 3$. Now consider the following sequence:

$$ y_n = \sum_{k=3}^{\infty}{\frac{1}{x_k^n}}$$

Because the terms of the series $y_n$ decrease as $n$ increases, it suffices to prove than $y_2$ converges to show that all $y_n$ converge if $n>1$. $\zeta(2)$ is a known convergent series:

$$ \frac{x_k^2}{k^2}=\frac{k^2-2kH_k+H_k^2}{k^2}=\frac{1-2\frac{H_k}{k}+\left(\frac{H_k}{k}\right)^2}{1}\to1$$

By the limit comparison test, $y_2$ converges. Therefore all $y_n$ with $n>1$ converge and $\{y_n\}_{n=2}^\infty$ is a well defined sequence. A final sequence shall be defined:

$$z_n = \sqrt[n]{y_n}$$

$\sqrt[n]{a}\to 1$ as $n\to\infty$ for some constant positive real $a$. From the facts that $y_n$ is decreasing and $\sqrt[n]{y_2} \geq \sqrt[n]{y_n} \geq 0$ we can deduce that $\lim_{n\to\infty} z_n$ exists and is in $[0, 1]$.

Questions: Is my identity correct? How can this be shown? Is the above proof correct?

Bonus: From the previous result and the Cauchy root test we can deduce the following series converges. Can you find out more about its value?

$$ \sum_{n=2}^\infty{y_n}$$

Homeward
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  • I don't have an easy way to do this, but have you seen if this result appears to hold if you replace $H_k$ with part of the asymptotic expansion of the harmonic series? e.g. what if you replace $H_k$ with $\log k + \gamma$ or $\log k + \gamma + \frac{1}{2n}$ -- does it still appear to converge to 6/7? If so, I might try proving that that converges and then proving that both converge to the same thing. – Daniel McLaury Oct 17 '14 at 23:45
  • $\log k + \gamma + \frac{1}{2k}$, rather – Daniel McLaury Oct 17 '14 at 23:45
  • Replacing $H_k$ with $\log k + \gamma$ appears to change the result. I have been notified that $\sqrt[n]{\sum x_k^n}$ will give out the maximum of $x_k$ as $n\to\infty$ (which does seem to happen here), but I do not know how to show this. In addition, the bonus question still stands. – Homeward Oct 17 '14 at 23:53
  • Replacing $H_k$ with $\log k + \gamma + \frac{1}{2k}$ does not seem to preserve the result of $6/7$ either. – Homeward Oct 17 '14 at 23:55
  • Oh, I didn't even read your whole expression. Yeah, that's an $\ell^\infty$ norm, so you're just maximizing. That's explained here for instance:

    http://math.stackexchange.com/questions/326172/the-l-infty-norm-is-equal-to-the-limit-of-the-lp-norms

     – Daniel McLaury Oct 18 '14 at 02:20

1 Answers1

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Notice that $$\sqrt[p]{\sum_{k=3}^{\infty} (k - H_k)^{-p}}$$ is simply the $\ell^p$-norm of the sequence $$\frac{1}{3-H_3}, \frac{1}{4-H_4}, \ldots, \frac{1}{k-H_k}, \ldots$$ Taking the limit as $p\to \infty$, assuming it exists, gives us the $\ell^\infty$-norm of this sequence. (See The $ l^{\infty} $-norm is equal to the limit of the $ l^{p} $-norms. for a proof.)

However, this is a decreasing sequence of positive numbers, and the first term is $$\frac{1}{3 - (1 + \frac{1}{2} + \frac{1}{3})} = \frac{6}{7}.$$

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Daniel McLaury
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