Prove that if $\sum u_n$ is convergent, then $\sum\sqrt[n]n\,u_n$ is also convergent 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
If $u_n \geq 0$ for all sufficiently large $n$, yes. And this is easy. 

Proof: without loss of generality, assume $u_n &gt; 0$ 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. $\square$
$$\frac{a_{n+1}}{a_n}=\frac{(n+1)^{\frac{1}{n+1}} u_{n+1}}{(n)^{\frac{1}{n}} u_{n}}$$
$$\frac{a_{n+1}}{a_n}\le\frac{(n+1)^{\frac{1}{n}} u_{n+1}}{(n)^{\frac{1}{n}} u_{n}} $$ 
$$\frac{a_{n+1}}{a_n}\le\frac{(\frac{n+1}{n})^{\frac{1}{n}} u_{n+1}}{ u_{n}} $$ 
$$\frac{a_{n+1}}{a_n}\le\frac{(1+\frac{1}{n})^{\frac{1}{n}} u_{n+1}}{ u_{n}} $$ 

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