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!
 A: By the Cauchy Condensation Test:
$$\sum_{n=3}^{\infty} {\frac{1}{n \cdot \ln(n) \cdot \ln(\ln(n))^{\alpha}}} \text{ converges} \iff $$
$$\sum_{n=3}^{\infty} {\frac{2^n}{2^n \cdot \ln(2^n) \cdot \ln(\ln(2^n))^{\alpha}}} =
\sum_{n=3}^{\infty} {\frac{1}{n \ln(2) \cdot \ln(n\ln(2))^{\alpha}}} \text{ converges} \iff $$
$$\sum_{n=3}^{\infty} {\frac{1}{(n\ln(2)+\ln(\ln(2)))^{\alpha}}} \text{ converges} \iff \alpha>1$$
A: Note that $$\frac{1}{1- \alpha}\frac{d}{dx}(\log\log(x))^{1-\alpha}=(\log\log(x))^{-\alpha}\frac{1}{x\log x}$$
since $$(\log\log x)^\prime =\frac{1}{x\log x}$$
A: 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.
