Integral test for convergence

Question

Show that $$\sum_{n=3}^\infty \frac1{n \log n(\log \log n)^p}$$ converges if and only if $ p > 1$.

I know the hint for this: we have to use integral test so we know that the sum of $f(n)$ converges $\iff$ integral of $f(x)$ converges.

Anyone know how exactly to do this?

See main theorem and to evaluate the integral use substitution $u=\ln(\ln(x))$ and the problem will fall apart. — Mhenni Benghorbal, May 29 '13 at 17:29

Norbert · Answer 1 · 2013-05-30T05:36:30.733

1

Note that $$ \int\limits_3^{+\infty}\frac{dx}{x\log x(\log\log x)^p} =\int\limits_3^{+\infty}\frac{d(\log x)}{\log x(\log\log x)^p} =\int\limits_3^{+\infty}\frac{d(\log\log x)}{(\log\log x)^p} =\frac{(\log\log x)^{-p-1}}{-p-1}\Biggl|_3^{+\infty} =\lim\limits_{x\to+\infty}\frac{(\log\log x)^{-p-1}}{-p-1}-\frac{(\log\log 3)^{-p-1}}{-p-1}=+\infty $$ Hence by integral test the series $$ \sum\limits_{n=3}^{+\infty}\frac{1}{n\log n(\log\log n)^p} $$ diverges

edited May 30 '13 at 05:36

answered May 29 '13 at 13:50

Norbert

58,398

sorry, can you explain how you get that? – Risa May 29 '13 at 14:26
Is it clear that $d(\log x)=\frac{dx}{x}$ ? – Norbert May 29 '13 at 14:30
i think u mean f(x) = ln x, f'(x) = 1/x ..? – Risa May 29 '13 at 14:42
I think my biggest problem here is i dont know how to integrate 1/x log x (log log x)^p – Risa May 29 '13 at 14:50
@yes, since you correctly understod what I menat. In fact my hint is a way of integration - just put $\int$ sign in front of each fraction – Norbert May 29 '13 at 14:52
but how do i integrate for (log log x)^p ? – Risa May 29 '13 at 14:55
make change of variables $u=\log\log x$. Which expression will yo get then? – Norbert May 29 '13 at 14:56
i know ∫ 1/x ln x dx = ln(ln x) but then how to combine with the variables u = log log x ? – Risa May 29 '13 at 14:59
@Risa answer my last question, otherwise I will not help you – Norbert May 29 '13 at 15:08
i'm guessing its log x. sorry, i'm not very confident in this topic – Risa May 29 '13 at 15:16
No, after change $u=\log\log x$ you will get $\int du/u^p$. Is it clear? – Norbert May 29 '13 at 15:18
oh! yes i got that now. so ∫ 1/x log x (log log x)^p is it [ln (ln x)]*[u^(-p-1)]/[-p-1]? – Risa May 29 '13 at 15:25
No it is $\frac{u^{-p-1}}{-p-1}$ where $u=\log\log x$ – Norbert May 29 '13 at 15:27
yeah i got that too, but i meant after getting that, now as a whole i got [ln(ln x)u^(-p-1)]/[-p-1]. do i have to substitute u = log log x back into this expression? – Risa May 29 '13 at 15:31
No this is wrong, why did you want to multiply by $\log\log x$? It is redundat here – Norbert May 29 '13 at 15:40
cause that ln(ln x) is the integral of 1/x ln x. As I need to integrate the function then I can find the limit and then converges.. – Risa May 29 '13 at 15:44
@Risa you need to repeat integrals from the very begining... I gave up. Here is a complete solution $$\int\frac{dx}{x\log x(\log\log x)^p}=\int\frac{d(\log x)}{\log x(\log\log x)^p}=\int\frac{d(\log\log x)}{(\log\log x)^p}=\int \frac{du}{^p}=\frac{u^{-p-1}}{-p-1}=\frac{(\log\log x)^{-p-1}}{-p-1}$$ – Norbert May 29 '13 at 15:54
thanks for the effort thou i still dont understand. :( – Risa May 29 '13 at 16:07
@Risa, see my edits – Norbert May 30 '13 at 05:36
@Risa see my ans. i hope u understand. – Siddhant Trivedi May 30 '13 at 06:21

Siddhant Trivedi · Answer 2 · 2013-05-30T06:26:01.290

Using Cauchy condensation test,

For $p=1$,

Here $$a_n=\frac{1}{n\log n \log (\log n)}$$

By Cauchy condensation test $\sum a_n $& $\sum 2^n a_{2^n}$ converges or diverges together,

\begin{align*} \sum 2^n a_{2^n}&=\sum 2^n \frac{1}{2^n \log {2^n} \log (\log 2^n)}\\ &=\sum \frac{1}{n\log 2 \log (n \log 2)} \end{align*}

Since,

$\log 2 < 1$

$n \log 2 < n$

$\displaystyle \frac {1}{n \log 2} > \frac {1}{n}$

$\displaystyle \frac{1}{\underbrace{n \log 2}_{m} \log\underbrace{(n \log 2)}_{m}} > \frac{1}{n \log n}$

Check the behavior $\displaystyle \sum \frac{1}{m \log m} $ it is divergent for p=1,

There is one example which you have to solve,

$$\sum_{n=2}^\infty \frac{1}{n(\log n)^p} \ \ \text{if} \ \ p>0 $$

For $P>1$ converges and $P\leq 1$ diverges.

$\displaystyle \therefore \sum 2^n a_{2^n}$ is divergent

$\displaystyle \therefore \sum_{n=3}^\infty \frac{1}{n \log n \log (\log n)} $ is divergent.

I hope you can conclude behavior for $P>1$.

Integral test for convergence

2 Answers2