Vtyjkyuk

If $sum u_n$ is a convergent series then can it be concluded that
$sumsqrt[n]n,u_n$ 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 $u_n > 0$ for all $n$.
  $
  lim_nto infty sqrt[n]n = 1
  $, so $sqrt[n]nu_n sim_ntoinftyu_n$ 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 $$sqrt[n]n = e^fracln nn = 1+ fracln nn+Oleft(fracln^2 nn^2right)tag1$$
  when $nto infty$, and $lim_ntoinftyu_n = 0$ we have
  $$
  sqrt[n]nu_n = u_n + fracln nnu_n + varepsilon_n tag2.
  $$
  with $varepsilon_n = oleft(fracln^2 nn^2right)$.
  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
  $$
  sum_n sqrt[n]nu_n
  $$
  converges if, and only if,
  $$
  sum_n fracln nnu_n
  $$
  does. So it suffices to prove the latter: this is where Dirichlet's test comes in handy: we apply it with $a_n = fraclnn$, $b_n=u_n$, 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. $square$

$$fraca_n+1a_n=frac(n+1)^frac1n+1 u_n+1(n)^frac1n u_n$$
$$fraca_n+1a_nlefrac(n+1)^frac1n u_n+1(n)^frac1n u_n $$
$$fraca_n+1a_nlefrac(fracn+1n)^frac1n u_n+1 u_n $$
$$fraca_n+1a_nlefrac(1+frac1n)^frac1n u_n+1 u_n $$

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