# Show the series $\sum \frac{\log n}{n^2}$ converges by comparison

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.

$$\frac{\log n}{n^2}\le\frac{n^{1/2}}{n^2}=\frac1{n^{3/2}}$$

Because for any $\;\epsilon>0\;$, we have that for all but a finite number of values of $$\;n\in\Bbb N\;,\;\;\;\log n<n^\epsilon$$

• 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. – Clement C. Sep 8 '16 at 15:32
• @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 – DonAntonio Sep 8 '16 at 17:42
• @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. – qbert Sep 8 '16 at 19:15
• Isn't log n smaller than square root of n, for all n>0. – jnyan Sep 9 '16 at 4:10
• @jnyan Yes, it is. – DonAntonio Sep 9 '16 at 10:24

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.

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: $$\sum_{n\geq 1}\frac{\log n}{n^2}\color{red}{\leq} \sum_{n\geq 1}\frac{H_n}{n^2}-\gamma\zeta(2)-\sum_{n\geq 1}\frac{1}{2n^2(n+1)}=\color{red}{\frac{6+24\,\zeta(3)-\pi^2(1+2\gamma)}{12}}\tag{2}$$

We can also use the Cauchy (Schlömilch) condensation test and then the comparison test. $$\sum\limits_{k=1}^{\infty} \frac{\mathrm{e}^{k}\mathrm{ln}(\mathrm{e}^{k})}{\mathrm{e}^{2k}} = \sum\limits_{k=1}^{\infty} \frac{k}{\mathrm{e}^{k}}$$

We then compare this series with the convergent series $$\sum\limits_{k=1}^{\infty} \frac{k}{k^{3}}$$ and thus our original series converges.