Proof for, that the given series converges if $\alpha \gt 1$ & diverges Otherwise.

Question

So the series we are interested in is,$$\sum _2^\infty \frac{1}{n(\log n)^\alpha}$$ this one. It is said that it converges if $\alpha \gt1$ & diverges if $\alpha \le 1.$
My attempt to Proof:
For the sake of brevity, It is assumed that this inequality $2\le n\le\infty$ holds true streching upto the end of the proof.

Now, For $\alpha\le1$,
the inequality $$n^\alpha\lt n(\log n)^\alpha$$ holds true which means $$\frac{1}{n(\log n)^\alpha}\lt \frac{1}{n^\alpha}$$ but the series $\sum_2^\infty\frac{1}{n^\alpha}$ is a divergent series for $\alpha\le1$, therefore $\sum_2^\infty\frac{1} {n(\log n)^\alpha}$ is divergent for $\alpha\le1.$
For $\alpha\gt1$, function $\frac{1} {n(\log n)^\alpha}$ is positive, continuous & non increasing as $n$ increases. Therefore our required series converges if the integral $\int_2^\infty\frac{dn} {n(\log n)^\alpha}$ converges. After simplifying the improper integral, we bound to get $\frac{(\log2)^{1-\alpha}}{\alpha-1}$. For $\alpha\gt1$ the sequence will look like this, $\frac{1}{\log 2},\frac{1}{2(\log 2)^2},\frac{1}{3(\log 2)^3},\frac{1}{4(\log 2)^4},\frac{1}{5(\log 2)^5},$ & so on. And it is a converging sequence (I guess it is approaching to 0). Which means $\sum_2^\infty\frac{1}{n(\log n)^\alpha}$ for $\alpha\gt1$ is convergent.

** Is my approach correct?** For proving 2nd part(i.e. for $\alpha\gt1$) I used $Theorem$ $IV$ results from Taylor & Mann's Advanced Calculus 3rd edition, 578 pages.