Does $\sum_{n=1}^{\infty} \frac{1}{n^{1 + 1/n}}$ converge? Does $$\sum_{n=1}^{\infty} \frac{1}{n^{1 + 1/n}}$$ converge? If yes, to where?
I searched this specific series but couldn't find a solution.
 A: Hint: $\dfrac{1}{n^{1+1/n}} = \frac{1}{n} \exp(-\frac{1}{n}\ln(n)) > \frac{1}{n} \exp(-1)$ as $n \to \infty$.
A: $2^n\ge n$ for $n\ge1$ is easy to show. Hence $2\ge n^{1/n}$, so
$${1\over n^{1+1/n}}\ge{1\over2n}$$
which implies $\sum{1\over n^{1+1/n}}\ge{1\over2}\sum{1\over n}$, which diverges.
A: The series does not converge.
Consider whether $n^{\frac 1n}\ge k$ for some constant $k$.  If $k = e$, we have
$$\frac 1n\ln n\ge 1\to \ln n \ge n$$
But for all $n\ge 1, \ln n\lt n$, so we have an upper bound on $n^\frac 1n$ which means that the original sum is asymptotic to $\sum_{n=1}^\infty \frac 1{n}$.
A: You can try the limit comparison theorem (more general form):
$$\frac{\frac1{n\sqrt[n]n}}{\frac1{n\log n}}=\frac{\log n}{\sqrt[n]n}\xrightarrow[n\to\infty]{}\infty$$
and then your series diverges since the other diverges.
For the other one: you can use Cauchy's Condensation test as the sequence is monotonic convergent to zero:
$$\frac{2^n}{2^n\log2^n}=\frac1{\log2}\frac1n$$
and the last one is just a constant multiple of the harmonic series.
