Determining if $\sum_{n =3}^{\infty} \frac{1}{n^{1/n}\cdot n\cdot \log(n)}$ diverges or converges.

Question

How would I show $\displaystyle\sum_{n = 3}^{\infty} \dfrac{1}{n^{\frac{1}{n}} \cdot n \cdot \log(n)}$ converges or diverges?

I want to say $n^{\frac{1}{n}} \cdot n \cdot \log(n) \cong n^{1 + \frac{1}{n}}$ converges but the $\dfrac{1}{n}$ doesn't help.

Answer 1

So $n^{1/n}<2$ by induction, so $\dfrac{1}{n^{1+1/n}\log n}>\dfrac{1}{2n\log n}$ and it is known that $\displaystyle\sum\dfrac{1}{n\log n}=\infty$.