# Convergence of $\sum_{n=1}^\infty \frac{1}{n(1+ln(n))^{\alpha}}$

I am asked to study the convergence of the following series: $$\sum_{n=1}^\infty \frac{1}{n(1+ln(n))^\alpha}$$

I have been told that the answer is that the aforementioned series is convergent when alpha is bigger than 1, yet I am at a loss at how to prove it.

I suppose I should apply the limit comparison test, but I do not know which other succession I should pick.

HINT: $\sum a_n$ converge iff $\sum 2^n a(2^n)$ converge (Known as the Cauchy Condensation Test)
So the problem is reduced to check the convergence of $$\sum \frac{2^n}{2^n(1+n\ln 2)^\alpha}=\sum \frac{1}{(1+n\ln 2)^\alpha}$$ Note $$\sum \frac{1}{(\ln 2+n\ln 2)^\alpha}\le\sum \frac{1}{(1+n\ln 2)^\alpha}\le\sum \frac{1}{(1+n)^\alpha}$$ $$\Rightarrow \frac{1}{(\ln 2)^\alpha}\sum \frac{1}{(1+n)^\alpha}\le\sum \frac{1}{(1+n\ln 2)^\alpha}\le\sum \frac{1}{(1+n)^\alpha}$$
$\sum 1/(1+n)^\alpha$ is the well-known $p$-series which converge if $\alpha >1$.(Not $\alpha \le 1$ as stated)
Comparing it to an analogous integral, $$\int_1^\infty \frac{dx}{x(1+\ln{x})^\alpha}$$ and with a substitution $u=\ln{x}$, you obtain $$\int_0^\infty \frac{du}{(1+u)^\alpha}$$ which converges for $\alpha>1$ (not less than 1 as the problem states).