$\lim_{n\to\infty} \sum\limits_{k=1}^n \frac{\ln k}{n} ( 1-\{\frac{n}{k} \} ) ( (1-\frac{k}{n} \{\frac{n}{k}\} )-\frac{1}{2})$ Can someone help me prove that the limit approaches zero ? I know it does, but I can't prove it.
$$\lim_{n\to\infty}\sum\limits_{k=1}^n\frac{\ln k}{n}\bigg(1-\bigg\{\frac{n}{k}\bigg\}\bigg)\bigg(\frac{1}{2}-\frac{k}{n}\bigg\{\frac{n}{k}\bigg\}\bigg)$$
where $\displaystyle\left\{\frac{n}{k}\right\}$ is the fractional part of $\displaystyle \frac{n}{k}$.
 A: This is quite an old question, and I guess OP already have an answer. For the future reference, however, I write down an answer.
The limit turns out to diverge to $+\infty$. Indeed, let $f : [0, 1] \to \Bbb{R}$ by
$$ f(x) = (1 - \{1/x\})(\tfrac{1}{2} - x\{1/x\}) \quad \text{and} \quad f(0) = 0. $$
It is easy to check that


*

*$0 \leq f(x) \leq \tfrac{1}{2}$,

*$f$ is Riemann-integrable, and hence

*$ n^{-1} \sum_{k=1}^{n} f(k/n) \to \int_{0}^{1} f(x) \, dx =: C > 0$ as $n \to \infty$.


Now fix any $m$. Then for all $n > m$, non-negativity of $f$ shows that
$$ \sum_{k=1}^{n} \frac{\log k}{n} f\left(\frac{k}{n}\right) \geq \sum_{k=m}^{n} \frac{\log m}{n} f\left(\frac{k}{n}\right).$$
Taking liminf as $n \to \infty$, we get
$$ \liminf_{n\to\infty} \sum_{k=1}^{n} \frac{\log k}{n} f\left(\frac{k}{n}\right) \geq C \log m $$
for any $m$ and therefore the limit diverges to $+\infty$. ////
In particular, it suggests that the original problem where $\log$ is replaced by von Mangoldt function $\Lambda$ requires some clever estimates on $\Lambda$ to produce a vanishing limit.
