Convergence of $\sum_{n=1}^{\infty}{\sqrt[n]{n}-1 \over n}$ I am trying to conclude about the convergence of
$$\sum_{n=1}^{\infty}{\sqrt[n]{n}-1 \over n}$$
If I take ${\sqrt[n]{{\sqrt[n]{n}-1 \over n}}}= {\sqrt[n]{\sqrt[n]{n}-1} \over \sqrt[n]{n}}$ this looks like it converges to $0$ but I am suspicious about the numerator. At the same time, other techniques I know like the ratio test, $n(\sqrt[n]{n}-1)$ and comparisons don't yield meaningful results. Any hints?
 A: Hint. One may notice that, as $n \to \infty$,
$$
\sqrt[n]{n}=n^{1/n}=e^{\frac{\ln n}n}=1+\frac{\ln n}n+\mathcal O\left(\frac{\ln^2 n}{n^2}\right)
$$ giving
$$
\frac{\sqrt[n]{n}-1}n=\frac{\ln n}{n^2}+\mathcal O\left(\frac{\ln^2 n}{n^3}\right)
$$ and thus the convergence of the given series using Bertrand's result.
A: Since $n=\sqrt n\cdot\sqrt n\cdot1\cdot1\cdots1$, the Arithmetic-Geometric Mean inequality tells us
$$\sqrt[n]n\le{\sqrt n+\sqrt n+1+1+\cdots+1\over n}={2\sqrt n+(n-2)\over n}\lt1+{2\over\sqrt n}$$
It follows that
$$\sum_{n=1}^\infty{\sqrt[n]n-1\over n}\lt\sum_{n=1}^\infty{2\over n^{3/2}}$$
A: 
Applying the integral test, we have
$$\begin{align}
\int_1^\infty \frac{x^{1/x}-1}{x}\,dx&=\int_1^\infty \frac{e^{\frac1x \log(x)}-1}{x}\,dx \tag 1\\\\
&\le \frac{e}{e-1} \int_1^\infty \frac{\log(x)}{x^2}\,dx \tag 2\\\\
&=\frac{e}{e-1}
\end{align}$$
Therefore the series of interest converges!


NOTE:
In going from $(1)$ to $(2)$, we used the inequality $e^x\le \frac{1}{1-x}$ for $x<1$, which I established in THIS ANSWER using only the limit definition of the exponential function and Bernoulli's Inequality.  
We also used the fact that $1-\frac{\log(x)}{x}\ge \left(1-\frac1e\right)$ for $x\ge 1$.
Finally, we evaluated the integral $\int_1^\infty \frac{\log(x)}{x^2}\,dx=1$ by enforcing the substitution $x=e^{t}$ and using integration by parts to evaluate the resulting integral $\int_0^\infty te^{-t}\,dt=1$.
