How to test if $\sum_{n=1}^{\infty }\frac{1}{n^{1+\frac{1}{n}}}$ converges? How to test whether $\sum_{n=1}^{\infty }\frac{1}{n^{1+\frac{1}{n}}}$ diverges or converges? Can I apply ratio test? Thanks in advance.
 A: Note that we have $n^{1/n}=e^{\frac1n \log(n)}\le \frac1{1-\frac1n\log(n)}$ for $n\ge 1$.   Therefore, for 
$$\frac{1}{n^{1+1/n}}\ge \frac{1-\frac1n\log(n)}{n}\ge \frac{1-\log(3)/3}{n}$$
Comparison to the harmonic series shows that the series diverges.
A: I might use a limit comparison test. You can compare with $\frac{1}{n}.$ Here's a hint on how to do that:
If $ L = \lim_{n \to \infty}\frac{1/n}{\frac{1}{n^{1 + \frac{1}{n}}}} = n^{\frac{1}{n}}$. Take logarithms of both sides of this equation, use L'Hopital, and don't forget to solve for $L$ and not $ln(L)$!
A: Note $\frac{\frac{1}{n^{1 + \frac 1n}}}{\frac 1n} = $$n^{-\frac 1n} = e^{-\frac{\log n}{n}} \xrightarrow[n\to+\infty]{} 1$
hence the series asymptotically grows like $\sum \frac 1n$. By the limit comparison test, we conclude divergence. 
A: The Cauchy Condensation Test: If $(x_n)_n$ is monotonic for all sufficiently large $n$ then $\sum_nA(n)$ converges iff $\sum_n 2^nA(2^n)$ converges. 
Let $f(x)=\ln (x^{1+1/x})=(1+1/x)\ln x.$ Then $f'(x)=(1/x+1/x^2)+(-1/x^2)\ln x=(1/x)(1-(\ln x)/x))>0$. 
So $(f(n))_n$ is increasing. So $(1/\exp (f(n)))_n=(1/n^{(1+1/n}))_n=(A(n))_n$ is decreasing. 
We have $2^nA(2^n)=2^n/((2^n)^{(1+2^{-n})})=2^{-n2^{-n}}.$ Now $-n2^{-n}\to 0$ as $n\to \infty.$ So $2^nA(2^n)\to 1$ as $n\to \infty.$ So $\sum_n2^nA(2^n)$ diverges.
