Is $\sum\limits_{n=1}^{\infty}\frac{1}{n^k+1}=\frac{1}{2} $ for $k \to \infty$? This series  :$\displaystyle\sum_{n=1}^{\infty}\frac{1}{n^k+1}$ is convergent for every $k>1$ , it's seems that it has a closed form for every $k >1$, some calculations here in wolfram alpha show to me that the sum approach to $\frac{1}{2}$ for large $k$ , My question here is :

Question:
Does $\displaystyle\sum_{n=1}^{\infty}\frac{1}{n^k+1}\to\frac{1}{2} $ for $k \to \infty$?


Yes, indeed
$$
0\le \lim_{k\to \infty}\sum_{n\ge 2}\frac{1}{n^k+1}\le \lim_{k\to \infty} \int_1^\infty \frac{1}{x^k+1}\mathrm{d}x=\int_1^\infty \lim_{k\to \infty}\frac{1}{x^k+1}\mathrm{d}x=0
$$
by the dominated convergence.
