The key idea to obtain the below sum is to evaluate this integral :$\int_{0}^{\infty} \frac{dx}{(nx+k)^{n}}$ with $k$ is a positive integer this integral after evaluation gives to me :$$\int_{0}^{\infty} \frac{dx}{(nx+k)^{n}}=\frac{k^{1-n}}{n²-n }\tag{1}$$ , By the summation of RHS and LHS OF $(1)$ from $k=1$ to $n$ we get the following sum : $$\sum_{k=1}^{n}\frac{k^{1-n}}{n(n-1)}\tag{2}$$ , Wolfram alpha assumed that sum has a closed form as shown below , Now my question here according to the approximation below depend for some values of $n$ or any evaluation of $(2)$ :Is : $$\sum_{k=1}^{n}\frac{k^{1-n}}{n(n-1)}=0 $$ for $n\to \infty$ ?
When $n>3$ $$ \sum_{k=1}^{n}{\frac{k^{1-n}}{n(n-1)}}= \frac{1}{n(n-1)}\sum_{k=1}^{n}{\frac{1}{k^{n-1}}}< \frac{1}{n(n-1)}\sum_{k=1}^{\infty}{\frac{1}{k^2}}\to 0 \text{ as }n\to\infty $$