Convergence of a particular summation

Question

Does summation $$\sum_{n=1}^\infty\frac{1}{n(\log n)^{a}}$$ converge if $a>1$?

Answer 1

Hint: The convergence property easily follows from Cauchy Condensation test .

Edit: In more detail, the suggested series has the same convergence properties as

$$ \sum\limits_{n=1}^{\infty}\dfrac{2^n}{2^n\left(\log2^n\right)^a}{}={}\sum\limits_{n=1}^{\infty}\dfrac{1}{n^a\left(\log2\right)^a}{}={}\dfrac{1}{\left(\log2\right)^a}\sum\limits_{n=1}^{\infty}\dfrac{1}{n^a} $$

which, being a $p$-series, converges for $a>1$.

Answer 2

The integral test will also do the job easily.

$$\int_2^\infty {dt \over t \log(t)^a} = \int_{\log(2)}^\infty {dt \over t^a}.$$

Do not use $n = 1$ or you will have a domain annoyance.