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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.

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$$0\leq\frac{1}{n}\sum_{k=1}^{n}\frac{-\log\left(\frac{k}{n}\right)}{k^{1+1/n}}\leq\frac{1}{n}\sum_{k=1}^{n}-\log\left(\frac{k}{n}\right)\xrightarrow{n\to +\infty}\int_{0}^{1}-\log(x)\,dx = 1 $$ 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: $$ \frac{1}{n}\int_{n}^{+\infty}\frac{\log\left(\frac{x}{n}\right)}{x^{1+1/n}}\,dx = n^{1-\frac{1}{n}}.$$

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  • $\begingroup$ In the last part alternatively can we write$$ \frac{1}{n}\sum\limits_{k>n}\frac{\log\left(\frac{k}{n}\right)}{k^{1+1/n}}=\frac {1}{n^{1+1/n}}\frac{1}{n}\sum\limits_{k>n}\frac{\log\left(\frac{k}{n}\right)}{(k/n)^{1+1/n}}=\frac{1}{n^{1+1/n}}\int\limits_1^{\infty}\frac{\ln x}{x^{1+1/n}}dx$$ $\endgroup$ – guest May 12 '16 at 16:42
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    $\begingroup$ @guest: sure, you may deal with the $k>n$ part through Riemann sums, too. The outcome is just the same as before, since $$\int_{1}^{+\infty}\frac{\log x}{x^{1+1/n}}\,dx = n^2.$$ $\endgroup$ – Jack D'Aurizio May 12 '16 at 16:45
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    $\begingroup$ On the other hand, it is not surprising that $\zeta'(s)$ behaves like $-\frac{1}{(1-s)^2}$ in a right neighbourhood of $s=1$, since $\zeta(s)$ has a simple pole with residue $1$ at $s=1$. $\endgroup$ – Jack D'Aurizio May 12 '16 at 20:09
  • $\begingroup$ Thanks for your consideration. It would be useful if the expression were bounded. Now I am looking for a positive sequence $(a_n)\to 0$ such that $$a_n\sum\limits_{k=1}^{\infty}\frac{|(\ln n)^2-(\ln k)^2|}{k^{1+a_n}}$$ is convergent or bounded. $\endgroup$ – guest May 12 '16 at 20:30
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    $\begingroup$ @guest: so you are considering $\zeta''(s)$, that by the same argument as above, behaves like $\frac{2}{(1-s)^3}$ in a right neighbourhood of $s=1$. It follows that no such sequence may exist, but for any $\{a_n\}_{n\geq 0}$ converging to zero $$ a_n^{\color{red}3} \sum_{k=1}^{+\infty}\frac{|\log^2 n-\log^2 k|}{k^{1+a_n}}$$ is bounded. $\endgroup$ – Jack D'Aurizio May 12 '16 at 21:12

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