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 $k \to \infty$?

 A: Let
$$S_k = \sum_{n=2}^{\infty}\frac{1}{n^k}.$$
We know $S_2 <\infty.$ Note that for $k>2,$
$$S_k = \sum_{n=2}^{\infty}\frac{1}{n^{k-2}}\frac{1}{n^2}\le \sum_{n=2}^{\infty}\frac{1}{2^{k-2}}\frac{1}{n^2}= \frac{1}{2^{k-2}}S_2.$$
This implies $S_k \to 0$ (in fact at an exponential rate). Thus
$$\frac{1}{2}< \sum_{n=1}^{\infty}\frac{1}{n^k+1}= \frac{1}{2}+ \sum_{n=2}^{\infty}\frac{1}{n^k+1} < \frac{1}{2} + S_k,$$
and we're done by the squeeze theorem.
A: 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.
A: 
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!
A: 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.
