# How to evaluate this limit and its convergence? $\sum_{n=1}^\infty\frac{1}{n\sqrt[n]{n}}$

How to evaluate this limit $$\sum_{n=1}^\infty\frac{1}{n\sqrt[n]{n}}$$ and its convergence?

I tried ratio test, root test, Raabe's test. However, I'm not getting anywhere. Can you please help me? Thank you

• For $n > 4$, $\dfrac1{n\sqrt[n]{n}} > \dfrac1{n\log\,n}$. It can be shown through other means that $\sum\limits_{n=2}^\infty \frac1{n\log\,n}$ is divergent, so... Jan 10 '12 at 15:45
• @J.M. I wonder why you prefer $\sqrt[n]{n}\lt\log n$ for every $n\gt4$ to $\sqrt[n]{n}\lt2$ for every $n\geqslant1$...
– Did
Jan 10 '12 at 18:43
• Does this answer your question? Does the series $\sum\limits_{n=1}^\infty \frac{1}{n\sqrt[n]{n}}$ converge? Oct 26 '20 at 9:46

For $n$ sufficiently large, $\root n\of n<2$; so, $${1\over n\,\root n\of n}>{1\over 2n}$$ for sufficiently large $n$.
Since the series $\sum\limits_{n=1}^\infty {1\over 2n}$ diverges (it is essentially the harmonic series), it follows from the Comparison test that the series $\sum\limits_{n=1}^\infty {1\over n\,\root n\of n}$ diverges.
• For $n$ sufficiently large... Read: For every $n\geqslant1$.
Hint: search an equivalent of $\sqrt[n]{n}$ as $n$ goes to $+\infty$ and use the Limit comparison test.