While working on a question, I wanted to find a limit in closed form as known numbers but I could not find a way to express it.
$$\alpha=\lim_{n\to\infty}[(\sum_{k=1}^n\frac{1}{n\ln(1+\frac{k^2}{n^2})})-\frac{{n \pi}^2}{6}]$$
$$\ln (x+1)=\frac{x}{1} -\frac{x^2}{2} + \frac{x^3}{3} - \frac{x^4}{4}+ ....=\sum \limits_{k=1}^{\infty} (-1)^{k+1} \frac{x^k}{k}$$ $$\frac{1}{n\ln(1+\frac{x^2}{n^2})}=\frac{1}{n(\frac{x^2}{n^2} - \frac{x^4}{2n^4} + \frac{x^6}{3n^6}+ \cdots)}=\frac{1}{\frac{x^2}{n}- \frac{x^4}{2n^3} + \frac{x^6}{3n^5}+ \cdots}=\frac{n}{x^2}(\frac{1}{1 - \frac{x^2}{2n^2} + \frac{x^4}{3n^4}+ \cdots)}=\frac{n}{x^2}.({1 + \frac{x^2}{2n^2} - \frac{x^4}{12n^4}+ \cdots)}$$
$$\alpha=\lim_{n\to\infty}[(\sum_{k=1}^n\frac{1}{n\ln(1+\frac{k^2}{n^2})})-\frac{{n \pi}^2}{6}]=\lim_{n\to\infty}[(\sum_{k=1}^n \frac{n}{k^2}.(1 + \frac{k^2}{2n^2} - \frac{k^4}{12n^4}+ \cdots) )-\frac{{n \pi}^2}{6}]=\lim_{n\to\infty}[(n\sum_{k=1}^n \frac{1}{k^2} + \frac{1}{2n}\sum_{k=1}^n1 - \frac{1}{12n^3}\sum_{k=1}^n k^2+ \cdots) -\frac{{n \pi}^2}{6}]= \frac{1}{2} - \frac{1}{36}+ \cdots$$
The results from above we can express the requested number as
$$\alpha=\sum_{k=1}^{\infty} \frac{1}{2k-1} \frac{1}{k!}\frac{d^k}{{dx}^k}(\frac{x}{\ln(1+x)})|_{x=0}$$
maybe that way can help to see the number in closed form
Thanks for answers
EDIT: @user8268 noticed good point.Maybe we can find the result via integral variable change. I want to share the transform from sum into integral representation.
We know that $$\frac{{\pi}^2}{6}=\lim_{n\to\infty}\sum_{k=1}^n\frac{1}{k^2}$$
Thus we can write that $$\alpha=\lim_{n\to\infty}[(\sum_{k=1}^n\frac{1}{n\ln(1+\frac{k^2}{n^2})})-n\lim_{n\to\infty}\sum_{k=1}^n\frac{1}{k^2}]$$
$$\alpha=\lim_{n\to\infty}[(\sum_{k=1}^n \frac{1}{n}(\frac{1}{\ln(1+\frac{k^2}{n^2})})-\lim_{n\to\infty}\sum_{k=1}^n \frac{1}{n}(\frac{1}{\frac{k^2}{n^2}})]$$
$$\alpha=\lim_{n\to\infty}[\sum_{k=1}^n \frac{1}{n}(\frac{1}{\ln(1+\frac{k^2}{n^2})}-\frac{1}{\frac{k^2}{n^2}})]$$
We can write a sum as integral because we know a formula that $$\int _0^x {f(t) dt}=\lim_{n\to\infty} \frac{x}{n}\sum \limits_{k=1}^n f(\frac{kx}{n})$$
Thus $$\alpha=\int _0^1 {(\frac{1}{\ln(1+t^2)}-\frac{1}{t^2}) dt}$$ Now I can focus on the variable change and try to evaluate the integral as known closed form numbers such as $\ln2, e,\gamma$, $\pi$ , $\Gamma(\frac{1}{4})$ etc (or their combinations if it is possible )
