I need to test for convergence $\sum_{n=1}^{\infty} \frac{1}{{n}^{\frac{n+1}{n}}} $. I can guess that it's probably a problem for the comparison test, although I have no idea what to compare it with. All my tries have failed.
Thank you for your help.
Note that for large enough $n$, $$n^{(n+1)/n}=n^{1+1/n}<2n$$ because $n^{1/n}$ converges to $1$ and is therefore eventually smaller than $2$. Therefore your terms are larger than $\frac{1}{2n}$.
Note that
$$\frac{1}{{n}^{\frac{n+1}{n}}}\sim\frac1n$$
then the given series diverges by limit comparison test with $\sum \frac1n$.
Using the ratio test$$a_n=\frac{1}{{n}^{\frac{n+1}{n}}}\implies \log(a_n)=-\frac{n+1}n \log(n)$$
$$\log(a_{n+1})-\log(a_n)=-\frac{n+2}{n+1} \log(n+1)+\frac{n+1}n \log(n)$$ Now, for large $n$, expand as a Taylor series to get $$\log(a_{n+1})-\log(a_n)=-\frac{1}{n}+\frac{\log \left({n}\right)-\frac{1}{2}}{n^2}+O\left(\frac{1}{n^3}\right)$$ $$\frac{a_{n+1}} {a_n }=e^{\log(a_{n+1})-\log(a_n) }=1-\frac{1}{n}+\frac{\log \left({n}\right)}{n^2}+O\left(\frac{1}{n^3}\right)$$
