TITLE: Proving a strange identity
QUESTION [5 upvotes]: 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?
$$
\sum_{n=2}^\infty{y_n}$$

REPLY [3 votes]: 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
$$\frac{1}{3 - (1 + \frac{1}{2} + \frac{1}{3})} = \frac{6}{7}.$$