Prove that if $\sum u_n$ is convergent, then $\sum\sqrt[n]n\,u_n$ is also convergent

Question

If $\sum u_n$ is a convergent series then can it be concluded that $\sum\sqrt[n]n\,u_n$ is also convergent

I think that by abels theorem since $\sqrt[n]n$ is monotone bounded hence the above theorem holds true

Please help from there

Answer 1

$$\frac{a_{n+1}}{a_n}=\frac{(n+1)^{\frac{1}{n+1}} u_{n+1}}{(n)^{\frac{1}{n}} u_{n}}$$ $$\frac{a_{n+1}}{a_n}\le\frac{(n+1)^{\frac{1}{n}} u_{n+1}}{(n)^{\frac{1}{n}} u_{n}} $$ $$\frac{a_{n+1}}{a_n}\le\frac{(\frac{n+1}{n})^{\frac{1}{n}} u_{n+1}}{ u_{n}} $$ $$\frac{a_{n+1}}{a_n}\le\frac{(1+\frac{1}{n})^{\frac{1}{n}} u_{n+1}}{ u_{n}} $$

$u_n$ is convergent also $(1+1/n)^{1/n}$

Answer 2

• If $u_n \geq 0$ for all sufficiently large $n$, yes. And this is easy.

  Proof: without loss of generality, assume $u_n > 0$ for all $n$. $ \lim_{n\to \infty} \sqrt[n]{n} = 1 $, so $\sqrt[n]{n}u_n \sim_{n\to\infty}u_n$ and you can conclude by the limit comparison test.

• If $(u_n)_n$ is allowed arbitrary signs, yes. But this is less obvious (though not horrendous either), and as you suggested follows from Abel/Dirichlet's test. (What follows is not the shortest proof, I reckon, but I find it instructive.) Note that since $$\sqrt[n]{n} = e^{\frac{\ln n}{n}} = 1+ \frac{\ln n}{n}+O\left(\frac{\ln^2 n}{n^2}\right)\tag{1}$$ when $n\to \infty$, and $\lim_{n\to\infty}{u_n} = 0$ we have $$ \sqrt[n]{n}u_n = u_n + \frac{\ln n}{n}u_n + \varepsilon_n \tag{2}. $$ with $\varepsilon_n = o\left(\frac{\ln^2 n}{n^2}\right)$. By assumption, $\sum_n u_n$ is a convergent series. By comparison with a $p$-series, the series $\sum_n \varepsilon_n$ is absolutely convergent. So overall, we have that $$ \sum_n \sqrt[n]{n}u_n $$ converges if, and only if, $$ \sum_n \frac{\ln n}{n}u_n $$ does. So it suffices to prove the latter: this is where Dirichlet's test comes in handy: we apply it with $a_n = \frac{\ln}{n}$, $b_n=u_n$, so that (i) $(a_n)_n$ is non-increasing with limit $0$; (ii) the partial sums of $\sum_n b_n$ are uniformly bounded (since $\sum_n u_n$ is convergent. $\square$