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 of course, we have the following:
$\frac{1}{2}\leq \sum\limits_{n=1}^\infty \frac{1}{n^k+1}\leq\frac{1}{2}+\int\limits_1^\infty \frac{1}{x^k+1}dx\leq \frac{1}{2}+ \int\limits_1^\infty \frac{1}{x^k}dx=\frac{1}{2}+\frac{1}{k+1}$.
Now use the theorem of squeezing.
