Study the converge $\sum_{n=1}^{\infty}(\sqrt[n]{n}-1)$ I need to study the convergence of the series 
$\sum_{n=1}^{\infty}(\sqrt[n]{n}-1)$
Now, I know that if we have a series $\sum_{n=1}^{\infty}a_n$ with positives elements and we can find a series $\sum_{n=1}^{\infty}b_n$ so that $0<a_n<b_n$ then if $\sum_{n=1}^{\infty}b_n$ is convergent then $\sum_{n=1}^{\infty}a_n$ is convergent.
Else, if $\sum_{n=1}^{\infty}a_n$ is divergent then $\sum_{n=1}^{\infty}b_n$ is divergent. The problem is I do not really know how to choose the series. Can you help me out?

The series diverges by comparison with $\sum \frac{1}{n}\log n$.
Let $n \ge 2$. Since $\sqrt[n]{n} = e^{\frac{1}{n}\log n}$, the mean value theorem gives $\sqrt[n]{n} - 1 = e^{c_n}\cdot \frac{1}{n}\log n$ for some $c_n\in \left(0, \frac{1}{n}\log n\right)$. Now $e^{c_n} > 1$, so that $$\sqrt[n]{n} - 1 > \frac{1}{n}\log n$$ Now you can compare your series with the divergent series $\sum \frac{1}{n}\log n$.
