I know that $$ \sum \frac{\log n}{n^2} $$ Converges by the integral test $$ \int_1^{\infty}\frac{\log x}{x^2}dx=-\frac{\log x}{x}+\int_1^{\infty}\frac{1}{x^2}dx=c-\lim_{x\rightarrow}\frac{\log x}{x}=c $$ But I was curious as to whether there was a clever way to show convergence by comparison.
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asked Sep 8 '16 at 15:07
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1$\begingroup$ $$\sum_{n\geq 1}\frac{\log n}{n^2}=-\zeta'(2) = \zeta(2)\left(\gamma+\log(2\pi)-12\log A\right)$$ where $A$ is the en.wikipedia.org/wiki/Glaisher%E2%80%93Kinkelin_constant $\endgroup$ – Jack D'Aurizio Sep 8 '16 at 15:11
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$\begingroup$ Possible duplicate of Convergence of $\frac{\ln(n)}{n^2}$ $\endgroup$ – Arnaud D. Dec 3 '19 at 11:08
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$$\frac{\log n}{n^2}\le\frac{n^{1/2}}{n^2}=\frac1{n^{3/2}}$$
Because for any $\;\epsilon>0\;$, we have that for all but a finite number of values of $$\;n\in\Bbb N\;,\;\;\;\log n<n^\epsilon$$
answered Sep 8 '16 at 15:08
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$\begingroup$ The "almost all values" is quite a bad phrasing: it could mean different things to different readers. "For all but a finite number" is much better; otherwise, one could consider the statement: "for almost all values of $n$, $n^2 \mathbb{1}_{n \rm is prime} = 0$" (which is true; the density of primes is going to $0$), but the series $\sum_n \frac{n^2 \mathbb{1}_{n \rm is prime}}{n^2}$ does not converge. $\endgroup$ – Clement C. Sep 8 '16 at 15:32
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1$\begingroup$ @ClementC. I agree with you, though "almost all" has, as far as I know, a pretty definite meaning in analysis in these things (as oposed as measure theory, say). Yet the OP probably hasn't yet studied that, so I shall change it. Thanks $\endgroup$ – DonAntonio Sep 8 '16 at 17:42
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1$\begingroup$ @DonAntonio Just fyi I have seen it, I just haven't seen much intro calc, particularly series. Although I imagine people in the future reading this may not. $\endgroup$ – operatorerror Sep 8 '16 at 19:15
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$\begingroup$ Isn't log n smaller than square root of n, for all n>0. $\endgroup$ – jnyan Sep 9 '16 at 4:10
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1$\begingroup$ For example, show (using L'Hospital's Rule or however) that $$\lim_{n\to\infty}\frac{\log n}{n^\epsilon}=0$$ for any $\;\epsilon>0\;$ ... $\endgroup$ – DonAntonio Jan 13 '20 at 9:31
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For large $n$, $\log n < n^{\frac{1}{2}}$. And,
$$ \sum\frac{n^{\frac{1}{2}}}{n^{2}}=\sum\frac{1}{n^{\frac{3}{2}}}<\infty, $$
by the $p$-test since $\frac{3}{2}>1$.
Therefore, the original series converges.
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An unusual proof may go through the following lines: since for any $n\geq 1$ we have $$ \log(n) \leq H_n-\gamma-\frac{1}{2(n+1)} \tag{1}$$ it follows that: $$ \sum_{n\geq 1}\frac{\log n}{n^2}\color{red}{\leq} \sum_{n\geq 1}\frac{H_n}{n^2}-\gamma\zeta(2)-\sum_{n\geq 1}\frac{1}{2n^2(n+1)}=\color{red}{\frac{6+24\,\zeta(3)-\pi^2(1+2\gamma)}{12}}\tag{2} $$
answered Sep 8 '16 at 18:15
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We can also use the Cauchy (Schlömilch) condensation test and then the comparison test. \begin{equation} \sum\limits_{k=1}^{\infty} \frac{\mathrm{e}^{k}\mathrm{ln}(\mathrm{e}^{k})}{\mathrm{e}^{2k}} = \sum\limits_{k=1}^{\infty} \frac{k}{\mathrm{e}^{k}} \end{equation}
We then compare this series with the convergent series $$\sum\limits_{k=1}^{\infty} \frac{k}{k^{3}}$$ and thus our original series converges.
