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 $k \to \infty$?



I thought it might be instructive to present an approach that relies on basic principles.  To that end we proceed.


Let $f(k)$ be the function given by $f(k)=\sum_{n=1}^\infty \frac{1}{1+n^k}$ for $k>1$.  Clearly, $f(k)=\frac12+\sum_{n=2}^\infty \frac{1}{1+n^k}$.  
We let $g(k)$ denote the series given by $g(k)=\sum_{n=2}^\infty \frac{1}{1+n^k}$ and note that $g(k)$ is monotonically decreasing.  
Let $\epsilon>0$ be given.  There exists a number $N'$ such that 
$$\sum_{n=N}^\infty \frac{1}{1+n^2}<\epsilon/2$$
whenever $N>N'$.  
We fix $N>N'$ and take $k$ so large that $\sum_{n=2}^{N-1}\frac{1}{1+n^k}< \frac\epsilon2$.  Then, we can write
$$\begin{align}
g(k)&=\sum_{n=2}^{N-1}\frac{1}{1+n^k}+\sum_{n=N}^\infty\frac{1}{1+n^k}\\\\
&\le \sum_{n=2}^{N-1}\frac{1}{1+n^k}+\sum_{n=N}^\infty\frac{1}{1+n^2}\\\\
&<\frac\epsilon2+\frac{\epsilon}{2}\\\\
&=\epsilon
\end{align}$$
We can assert, therefore, that 
$$\lim_{k\to \infty}g(k)=0$$
and hence we have
$$\lim_{k\to \infty}f(k)=\frac12$$
as was to be shown!
