Is the given expression convergent as $n\to\infty$?

Question

I want to know whether the following expression is convergent as $n\to\infty$ $$\frac{1}{n}\sum\limits_{k=1}^{\infty}\frac{|\ln n-\ln k|}{k^{(1+1/n)}}\cdot$$ With use of Riemann zeta function $\zeta(s)$ with $s=1+\frac{1}{n}$ and using the fact $\lim\limits_{s\to 1^+}\zeta(s)(s-1)=1$, the expression is equivalent to (as $n\to\infty$) $$\frac{1}{\zeta(1+1/n)}\sum\limits_{k=1}^{\infty}\frac{|\ln n-\ln k|}{k^{(1+1/n)}}\cdot$$ Only showing boundedness or unboundedness may also be useful. Thanks for your helps.

Answer 1

$$0\leq\frac{1}{n}\sum_{k=1}^{n}\frac{-\log\left(\frac{k}{n}\right)}{k^{1+1/n}}\leq\frac{1}{n}\sum_{k=1}^{n}-\log\left(\frac{k}{n}\right)\xrightarrow{n\to +\infty}\int_{0}^{1}-\log(x)\,dx = 1 $$ hence the relevant part is just: $$ \frac{1}{n}\sum_{k>n}\frac{\log\left(\frac{k}{n}\right)}{k^{1+1/n}} $$ that is diverging, since such a series behaves like: $$ \frac{1}{n}\int_{n}^{+\infty}\frac{\log\left(\frac{x}{n}\right)}{x^{1+1/n}}\,dx = n^{1-\frac{1}{n}}.$$