Convergence of $\sum_{n=2}^\infty \frac{1}{nn^{1/n}}$ I'm trying to find out if the series
$$\sum_{n=2}^\infty \frac{1}{nn^{1/n}}$$
converges or not. First with the ratio test and then with the integral test.

Ratio test:
$$r=\lim_{n\rightarrow \infty}\frac{a_{n+1}}{a_n}=\lim\frac{nn^{1/n}}{(1+n)(1+n)^{1/(n+1)}}$$
$$\lim_{n\rightarrow \infty}\frac{n}{n+1}=1$$
$$\lim_{n\rightarrow \infty}n^{1/n}=\lim e^{\ln(n)/n}=1$$
$$\lim_{n\rightarrow \infty}(n+1)^{1/(n+1)}=\lim e^{\ln(n+1)/(n+1)}=e^0=1$$
That means $r=1$ so ratio text is inconclusive.

Integral test
$$\int^L \frac{1}{xx^{1/x}}dx=\int^L\frac{1}{xe^{\ln(x)/x}}dx$$
Let $\ln(x)=y$ then
$$\int\frac{1}{\exp\left\{ye^{-y}\right\}}dy $$
I don't know, what do now? How to show if this integral will or will not converge as $L\rightarrow \infty$?

There are plenty of proofs that $2^n>n,$ for all integers $n\geq 0.$ Thus $0<n^{1/n}<2$ and thus:
$$\frac1{nn^{1/n}}>\frac{1}{2n}$$
