$\lim_{n\to\infty}\left(\sum_{k=0}^{n-1}(\zeta(2)-H_{k,2})-H_n\right)=1$

Question

I found this limit in a book, without any explanation:

$$\lim_{n\to\infty}\left(\sum_{k=0}^{n-1}(\zeta(2)-H_{k,2})-H_n\right)=1$$

where $H_{k,2}:=\sum_{j=1}^k\frac1{j^2}$. However Im unable to find the value of this limit from myself. After some work I get the equivalent expression

$$\lim_{n\to\infty}\sum_{k=0}^{n-1}\sum_{j=k}^\infty\frac1{(j+1)^2(j+2)}$$

but anyway Im stuck here. Can someone show me a way to compute this limit? Thank you.

UPDATE: Wolfram Mathematica computed it value perfectly, so I guess there is some integral or algebraic identity from where to calculate it.

Answer 1

Let's see:

$$\begin{eqnarray*} \sum_{k=0}^{n-1}\left(\zeta(2)-H_k^{(2)}\right) &=& \zeta(2)+\sum_{k=1}^{n-1}\left(\zeta(2)-H_k^{(2)}\right)\\&\stackrel{\text{SBP}}{=}&\zeta(2)+(n-1)(\zeta(2)-H_{n-1}^{(2)})+\sum_{k=1}^{n-2}\frac{k}{(k+1)^2}\\&=&\zeta(2)+(n-1)(\zeta(2)-H_{n-1}^{(2)})+(H_{n-1}-1)-\sum_{k=1}^{n-2}\frac{1}{(k+1)^2}\\&=&n( \zeta(2)-H_{n-1}^{(2)})+H_{n-1}\end{eqnarray*}$$ hence the claim is equivalent to

$$ \lim_{n\to +\infty} n(\zeta(2)-H_{n-1}^{(2)}) = \lim_{n\to +\infty}n\sum_{m\geq n}\frac{1}{m^2} = 1 $$ which is pretty clear since $\sum_{m\geq n}\frac{1}{m^2} = O\left(\frac{1}{n^2}\right)+\int_{n}^{+\infty}\frac{dx}{x^2}=\frac{1}{n}+O\left(\frac{1}{n^2}\right)$.
$\text{SBP}$ stands for Summation By Parts, of course.

Answer 2

Considering your last expression $$a_n=\sum_{k=0}^{n-1}\sum_{j=k}^\infty\frac1{(j+1)^2(j+2)}$$ $$\sum_{j=k}^\infty\frac1{(j+1)^2(j+2)}=\psi ^{(1)}(k+1)-\frac{1}{k+1}$$ making $$a_n=n \,\psi ^{(1)}(n+1)$$ the expansion of which being $$a_n=1-\frac{1}{2 n}+\frac{1}{6 n^2}+O\left(\frac{1}{n^4}\right)$$