# Infinite series of $\frac{1}{n^p+1}$

I was wondering if the infinite series $$\sum_{n=0}^\infty \frac{1}{n^{3/2}+1}\approx 1.95202$$ can be expressed in closed form (unlikely) or in terms of the fractional values of the zeta function (more likely). More generally, I would like to know if anyone can find a way to do so for $$\sum_{n=0}^\infty \frac{1}{n^{p}+1}$$ where $p\ne \mathbb Z$.

Wolfram yields nothing but the numerical approximation above.

$$\sum_{n=0}^\infty \frac{1}{n^{p}+1}=\frac{3}{2}+\sum_{n=2}^\infty \frac{1}{n^{p}+1}$$
The latter series can be expressed as a double series (we assume $p>0$):
$$\sum_{n=2}^\infty \frac{1}{n^{p}+1}=\sum_{n=2}^\infty \frac{1}{n^p} \sum_{k=0}^\infty (-1)^k \frac{1}{n^{pk}} =\sum_{k=0}^\infty (-1)^k \sum_{n=2}^\infty \frac{1}{n^{(k+1)p}}$$
$$\sum_{n=0}^\infty \frac{1}{n^{p}+1}=\frac{3}{2}+\sum_{k=0}^\infty (-1)^k \left(\zeta \left( (k+1)p \right)-1 \right)$$
Where indeed, fractional zeta values appear for fractional $p$.