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I want to check if $\sum\limits_{n=2}^{\infty}\frac{1}{n\ln^2(n)}$ is convergent by using the integral criteria.

Therefore I want to solve the integral $\int_{2}^{\infty}\frac{1}{x\ln^2(x)} dx$

First I tried to start only with the denominator and to use partial integration there, but I didn't succeeded, as the formula gets more complex with every duration and after using it for three times, I didn't had the feeling that it will end. (With every duration I got the term $x*ln(x)-x$ one time more multiplied inside the integral.)

Therefore I tried to split it first into $\int_{2}^{\infty}\frac{1}{x}\times \frac{1}{\ln^2(x)} dx$ and then to use partial integration on the whole term. But even with this I couldn't get to a result. I am also not sure about how to integrate $\ln^2(x)$. It doesn't seem to be correct to just integrate $\ln (x)$ and then take the square of the result. Also when I want to solve $\int \ln (x)\times\ln (x)dx$ first, by using partial integration, I encounter the same problem as before: I can't get to an end as the part under the integral grows with every iteration.

How can I solve the integral correctly to know if the sum converges? In this example I think that it should converge, but if I had a divergent series, how can I find out, if the integral exists or not? Just by trying to integrate it?

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  • $\begingroup$ would you like to have another routine method? $\endgroup$ – mrs Sep 21 '18 at 12:41
  • $\begingroup$ Yes, would be interesting. $\endgroup$ – mrs fourier Sep 21 '18 at 12:46
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Note that $\frac1x=\ln'x$. Therefore$$\int\frac1{x\ln^2x}\,\mathrm dx=\int\frac{\ln'x}{\ln^2x}\,\mathrm dx.$$So, if you do $\ln x=t$ and $\frac1x\,\mathrm dx=\mathrm dt$, you get$$\int\frac1{t^2}\,\mathrm dt=-\frac1t=-\frac1{\ln x}.$$

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  • $\begingroup$ Thanks for the reply, but I can't relieve why we are allowed to substitute 1/x dx with only dt, and without replacing it later again, as so a part of the formula gets lost. Also I can't understand how to solve 1/t² easily as it leads me to (ln t)/t-1/t*2*integral((t*ln t - t)/t³) dt after applying partial derivation on it for two times. $\endgroup$ – mrs fourier Sep 21 '18 at 14:12
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    $\begingroup$ When we do a substitution $t=f(x)$, we should also do $\mathrm dt=f'(x)\,\mathrm dx$. So, if we do $t=\ln x$, we also do $\mathrm dt=\frac1x\mathrm dx$. And, since $\left(\frac1t\right)'=-\frac1{t^2}$, $-\frac1t$ is a primitive of $\frac1{t^2}$. $\endgroup$ – José Carlos Santos Sep 21 '18 at 14:15
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You can indeed use integration by parts. As $(\ln x)' = \frac{1}{x}$ and $(\frac{1}{(\ln x)^2})' = -2\frac{1}{x(\ln x)^3}$ we have for $R \gt 0$ $$ \int_2^R \frac{1}{x(\ln x)^2} \,\mathrm{d}x = \left[\frac{1}{\ln x}\right]_2^R + 2 \int_2^R \frac{1}{x(\ln x)^2} \,\mathrm{d}x, $$ hence $$ \int_2^R \frac{1}{x(\ln x)^2} \,\mathrm{d}x - 2 \int_2^R \frac{1}{x(\ln x)^2} \,\mathrm{d}x = \frac{1}{\ln R} - \frac{1}{\ln 2} $$ and therefore $$ \int_2^R \frac{1}{x(\ln x)^2} \,\mathrm{d}x = \frac{1}{\ln 2} - \frac{1}{\ln R} \to \frac{1}{\ln 2} \lt \infty $$ for $R \to \infty$.

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  • $\begingroup$ Thanks, this makes it a bit clearer, but I still don't know, how it continues after the word "hence"? Why are we allowed to continue without solving the integral? (When I try to solve it by applying partial integration for another time, I again got the same term aggregated in an endless loop.) Why does the upper limit gets subtracted from the lower one, doesn't we always start with the upper limit and then subtract the lower one from it? Why only ln(x) is used instead of x(ln x)² when fillling in the both limits? $\endgroup$ – mrs fourier Sep 21 '18 at 13:44
  • $\begingroup$ I subtracted $2 \int \dots$ and multiplied with $-1$. Edit: I have added a step in the calculation above. $\endgroup$ – Keba Sep 21 '18 at 13:56
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Hint:

$$\int\frac{dx}{x\log^2x}=\int\frac{d(\log x)}{\log^2x}.$$

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