For which $\alpha$ this sum converges? $\sum_{n=3}^{\infty} {\frac{1}{n \cdot \ln(n) \cdot \ln(\ln(n))^{\alpha}}}$

Given:

$$\sum_{n=3}^{\infty} {\frac{1}{n \cdot \ln(n) \cdot \ln(\ln(n))^{\alpha}}}$$ I am asked: For what values of $\alpha$ does this summation converge?

So I said, $f(n) = \frac{1}{n \cdot \ln(n) \cdot \ln(\ln(n))^{\alpha}}$. $f(n)$ is obviously monotonically decreasing function. Then, by using the integral test, this summation converges if and only if $I = \int_{3}^{\infty} {\frac{dn}{n \cdot \ln(n) \cdot \ln(\ln(n))^{\alpha}}}$ converges. (has a value)

But I am finding this very hard to go on with. Any direction will be appreciated!

• The parenthesis should be for the outside $ln$ – Sungjin Kim May 29 '13 at 19:30
• Hint: $u=\ln\ln(n)$ – Zen May 29 '13 at 19:32
• Zen, please write your hint as an answer so I can accept it, it worked. we get to the integral $\int_{ln(ln(3))}^{\infty} {\frac{dt}{t^{\alpha}}}$ which is convergent for $\alpha > 1$ and divergent for $\alpha \leq 1$. – TheNotMe May 29 '13 at 20:32

By the Cauchy Condensation Test: $$\sum_{n=3}^{\infty} {\frac{1}{n \cdot \ln(n) \cdot \ln(\ln(n))^{\alpha}}} \text{ converges} \iff$$ $$\sum_{n=3}^{\infty} {\frac{2^n}{2^n \cdot \ln(2^n) \cdot \ln(\ln(2^n))^{\alpha}}} = \sum_{n=3}^{\infty} {\frac{1}{n \ln(2) \cdot \ln(n\ln(2))^{\alpha}}} \text{ converges} \iff$$ $$\sum_{n=3}^{\infty} {\frac{1}{(n\ln(2)+\ln(\ln(2)))^{\alpha}}} \text{ converges} \iff \alpha>1$$

• For other readers, because this took me a while to figure out: the second arrow is not a trivial rewrite but a second consecutive application of the Cauchy Condensation Test. – Myself May 29 '13 at 21:06

Note that $$\frac{1}{1- \alpha}\frac{d}{dx}(\log\log(x))^{1-\alpha}=(\log\log(x))^{-\alpha}\frac{1}{x\log x}$$

since $$(\log\log x)^\prime =\frac{1}{x\log x}$$

Peter's solution and mine are the same, but the idea is to make a u-substitution: $$\int_3^\infty \frac{dn}{n\ln n (\ln \ln n)^\alpha}$$

The cleanest substitution is $u=\ln\ln n$, since then $du=\frac{1}{\ln n}\cdot \frac{1}{n} dn$ which nicely ties up your denominator into $$\int_{\ln(\ln (3))}^\infty \frac{1}{u^\alpha} du$$ and the rest is obvious.

Even if you didn't see the substitution $u=\ln\ln n$, you could at least notice that if you subbed $u=\ln n$ you could get rid of the $1/n$ and be left integrating $\frac{1}{u\ln u}$, which you could apply the same technique to.