Summing a series to $n^{2}$

Question

I'm currently trying to to sum the following series:

$$ \sum_{k=1}^{n^2}\frac{1}{1 + \left ( \frac{k}{n} \right)^{r}} $$

I'm not sure what to do since we are summing over $n^{2}$. Any help would be appreciated.

Answer 1

Assuming that $r\gt0$, the $n$th sum $S_n$ is $n$ times a Riemann sum of the decreasing function $x\mapsto1/(1+x^r)$ on the interval $(0,n)$ hence $$n\int_0^n\frac{\mathrm dx}{1+x^r}-1+\frac1{1+n^r}\leqslant S_n\leqslant n\int_0^n\frac{\mathrm dx}{1+x^r}.$$ In particular, if $r\gt1$, then, when $n\to\infty$,$$\frac1{n}S_n\to\int_0^\infty\frac{\mathrm dx}{1+x^r}=\frac{\pi}{r\sin\left(\pi/r\right)},$$ if $r=1$, then $$\frac1{n\log n}S_n\to1,$$ and, if $0\lt r\lt1$, then $$\frac1{n^{2-r}}S_n\to\frac1{1-r}.$$

Answer 2

If $r$ is a positive integer, you can expand $1/(1+(k/n)^r)$ in partial fractions:

$$ \dfrac{1}{1+z^r} = \sum_{\alpha} \dfrac{-\alpha}{r (z - \alpha)}$$ the sum being over the $r$'th roots of $-1$.

For each such $r$ we get a "closed-form" expression in terms of the $\Psi$ function

$$ \sum_{k=1}^{n^2} \dfrac{1}{1+(k/n)^r} = \sum_\alpha \sum_{k=1}^{n^2} \dfrac{-\alpha}{r (k/n - \alpha)} = \sum_\alpha \dfrac{\alpha n}{r} \left( \Psi(1-\alpha n) - \Psi(1-\alpha n + n^2)\right) $$