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 $L\rightarrow \infty$?
 A: There are plenty of proofs that $2^n>n,$ for all integers $n\geq 0.$ Thus $0<n^{1/n}<2$ and thus:
$$\frac1{nn^{1/n}}>\frac{1}{2n}$$
A: Comparison with Harmonic Series
As shown in this answer
$$
n^{1/n}\le1+\sqrt{\frac2n}\tag1
$$
For $n=1$, $n^{1/n}=1$. Inequality $(1)$ shows that for $n\ge2$, we have $n^{1/n}\le2$. This means that for $n\ge1$, $n^{1/n}\le2$. Thus,
$$
\sum_{n=1}^\infty\frac1{nn^{1/n}}\ge\sum_{n=1}^\infty\frac1{2n}\tag2
$$
which diverges since the Harmonic Series diverges.

Cauchy Condensation Test
Since the terms tend monotonically to $0$,
$$
\sum_{n=1}^\infty\frac1{nn^{1/n}}\tag3
$$
converges if and only if
$$
\sum_{n=1}^\infty\frac{2^n}{2^n2^{n/2^n}}\tag4
$$
The terms of $(4)$ tend to $1$, so the series diverges by the Term Test.
A: As $k$ gets large, the power on $k,$ $1 + \frac1{k},$ approaches $1.$ So, it should make sense to do the limit comparison test with $\sum_{k = 1}^{\infty} \frac1{k}.$
So, consider the limit $\lim_{k \to \infty} \frac{\frac1{k}}{\frac{1}{kk^{1/k}}} = \lim_{k \to \infty} k^{\frac1{k}} = \lim_{k \to \infty} e^{\frac{\ln k}{k}}.$ Because $f(x) = e^x$ is continuous everywhere, we can say this is equal to $e^{\lim_{k \to \infty} \frac{\ln k}{k}} = e^0 = 1.$
Now, because the limit is finite, either both series converge or both series diverge. The harmonic series $\sum_{k = 1}^{\infty} \frac1{k}$ famously diverges, so our series also diverges.
