Evaluation of a series involving the harmonic number

Question

Let $\text{H}_n$ denote the n-th harmonic number, does the following sequence have a closed-form ? As an approximation I have got $0.0922514...$ $$ \frac{3}{2}+\lim_{n\to\infty} \bigg(\sum_{k=3}^{n}\frac{2}{k+\sqrt{k^2-4}}-\text{H}_n\bigg)$$

Answer 1

$$S(n)=\sum^n_{k=3}\frac{2}{{k+\sqrt{k^2-4}}}=\sum^n_{k=3}\frac{2}{{k+\sqrt{k^2-4}}}\cdot\frac{k-\sqrt{k^2-4}}{k-\sqrt{k^2-4}}=\frac12\sum^n_{k=3}k-\sqrt{k^2-4}$$ which can be simplified to $$\frac12\left(\frac{n(n+1)}2-3-{\sum^n_{k=3}\sqrt{k^2-4}}\right)$$

By generalized binomial theorem,

$$\begin{align} \sum^n_{k=3}\sqrt{k^2-4} & = \sum^n_{k=3}k\sqrt{1-\frac{4} {k^2}} \\ &=\sum^n_{k=3}k\sum^\infty_{m=0}\binom{1/2}{m}(-4)^m\frac1{k^{2m}} \\ &=\underbrace{\sum^n_{k=3}k}_{\text{when }m=0}+\underbrace{(-2)\sum^n_{k=3}\frac1k}_{\text{when }m=1}+\sum^n_{k=3}k\sum^\infty_{m=2}\binom{1/2}{m}(-4)^m\frac1{k^{2m}}\\ & =\frac{n(n+1)}2-2(H_n-\frac52)+\sum^n_{k=3}\sum^\infty_{m=2}\binom{1/2}{m}(-4)^m\frac1{k^{2m-1}}\\ \end{align} $$

$$S(n)=\frac12\left(-3+2H_n-5-\sum^n_{k=3}\sum^\infty_{m=2}\binom{1/2}{m}(-4)^m\frac1{k^{2m-1}}\right)$$

Therefore, $$\lim_{n\to\infty}S(n)-H_n=-4-\frac12\sum^\infty_{k=3}\sum^\infty_{m=2}\binom{1/2}{m}(-4)^m\frac1{k^{2m-1}}$$

You may try to express it in terms of zeta function.

EDIT: $$\binom{1/2}{m}(-4)^m=-\frac{C^{2m}_m}{2m-1}$$ So, $$\lim_{n\to\infty}S(n)-H_n=-4+\frac12\sum^\infty_{k=3}\sum^\infty_{m=2}\frac{C^{2m}_m}{(2m-1)k^{2m-1}}$$