I would like to show that
$\sum \frac{1}{nn^{1/n}}$ diverges, and I'm quite certain I will have to use the comparison test. I don't immediately see how that would be useful, though.
I would like to show that
$\sum \frac{1}{nn^{1/n}}$ diverges, and I'm quite certain I will have to use the comparison test. I don't immediately see how that would be useful, though.
$\displaystyle \sum \frac{2^n}{2^n(2^n)^{1/2^n}}>\sum \frac{2^n}{2^n(2^n)^{1/n}}=\sum\frac{1}{2}=∞$
HINT: Suppose that you can find a positive number $c$ and a positive integer $n_0$ such that $n^{1/n} for all $n\ge n_0$; then you’d have $\frac1{nn^{1/n}}\ge\frac1{cn}$ for all $n\ge n_0$. What do you know about $\lim_{n\to\infty}n^{1/n}\;?$
According to Cauchy's Condensation Test , our series converges iff the following series converges:
$\sum_{n=1}^\infty\frac{2^n}{2^n(2^n)^{1/2^n}}=\sum_{n=1}^\infty\frac{1}{2^{n/2^n}}$
But
$2^{n/2^n}=e^{\frac{n}{2^n}\log 2}\xrightarrow [n\to\infty]{}1\neq 0$
thus our series diverges.
Added: Another easy test: choose $\,\{b_n:=\frac{1}{n}\}\,$ ,then:
$\frac{\frac{1}{n\,n^{1/n}}}{b_n}=\frac{1}{\sqrt[n] n}\xrightarrow [n\to\infty]{}1\neq 0$
Thus, our series converges iff the series $\,\sum b_n\,$ does...which it doesn't as it is the harmonic series.