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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 $\displaystyle \frac{n}{k}$.

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Please consider adding $\LaTeX$ format to your question – Pragabhava Oct 26 '12 at 2:24
how do i do that – boby Oct 26 '12 at 2:26
@boby I typeset your fomulae using $\LaTeX$, please check if my interpretation is correct. Also to other members, is there a better notation for the Fractional function? – Frenzy Li Oct 26 '12 at 2:28
Thanks a bunch, could you help me prove it though? I tested it on wolfram alpha and it aproaches zero quite fast but I still cant prove it – boby Oct 26 '12 at 2:29
To answer your question "how do i do that":… – joriki Oct 26 '12 at 4:18

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.

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