Krdytkyu

I am studying the convergence of the series $$sumlimits_{n<=0} left(sqrt[n]{n} - 1right)^n$$
I just know that the limit tend to $0$, but I don't know how to prove the convergence.

I assume you meant $n geq 1$. Using $x^n-1=(x-1)(x^{n-1}+x^{n-2}+...+x^{2}+x+1)$ we have

$$n-1=
left(n^{frac{1}{n}}right)^{n}-1=\
left(n^{frac{1}{n}}-1right)left(n^{frac{1}{n}(n-1)}+n^{frac{1}{n}(n-2)}+...+n^{frac{1}{n}2}+n^{frac{1}{n}}+1right)geq ...$$
using AM-GM
$$... geqleft(n^{frac{1}{n}}-1right)left((n-1)sqrt[n-1]{n^{frac{1}{n}(n-1)}cdot n^{frac{1}{n}(n-2)}cdot ...cdot n^{frac{1}{n}2}cdot n^{frac{1}{n}}}+1right)=\
left(n^{frac{1}{n}}-1right)left((n-1)sqrt[n-1]{n^{frac{1}{n}(n-1+n-2+...+1)}}+1right)=
\
left(n^{frac{1}{n}}-1right)left((n-1)sqrt[n-1]{n^{frac{1}{n}frac{(n-1)n}{2}}}+1right)=\
left(n^{frac{1}{n}}-1right)left((n-1)sqrt{n}+1right)$$

Or
$$0< n^{frac{1}{n}}-1leq frac{n-1}{(n-1)sqrt{n}+1}<frac{1}{sqrt{n}} tag{1}$$

From $(1)$ we see that $limlimits_{nrightarrowinfty} left(n^{frac{1}{n}}-1right)=0$, which means

• 
  $0< left(n^{frac{1}{n}}-1right)^n<frac{1}{n^{frac{n}{2}}}$, thus $0<sumlimits_{ngeq1} left(n^{frac{1}{n}}-1right)^n leq sumlimits_{ngeq1} frac{1}{n^{frac{n}{2}}}$. The latter is converging by ratio test $$limlimits_{nrightarrowinfty}frac{frac{1}{(n+1)^{frac{n+1}{2}}}}{frac{1}{n^{frac{n}{2}}}}=
  limlimits_{nrightarrowinfty}frac{n^{frac{n}{2}}}{(n+1)^{frac{n+1}{2}}}=
  limlimits_{nrightarrowinfty}frac{1}{left(1+frac{1}{n}right)^{frac{n}{2}}cdot sqrt{n+1}}=frac{1}{sqrt{e}cdot limlimits_{nrightarrowinfty}sqrt{n+1}}=0$$
  


• or simpler version from the definition of the limit, for $forall n> N(varepsilon) Rightarrow 0<left(n^{frac{1}{n}}-1right)<varepsilon <1$, thus $0<left(n^{frac{1}{n}}-1right)^n<varepsilon^n$ and $$0<sumlimits_{ngeq1} left(n^{frac{1}{n}}-1right)^n <
  sumlimits_{n=1}^{N(varepsilon)} left(n^{frac{1}{n}}-1right)^n + sumlimits_{n=N(varepsilon)+1} varepsilon^n=
  sumlimits_{n=1}^{N(varepsilon)} left(n^{frac{1}{n}}-1right)^n + frac{e^{N(varepsilon)+1}}{1-varepsilon}$$
  we just need to find the very first $N(varepsilon)$ for $0<varepsilon<1$ and $sumlimits_{n=1}^{N(varepsilon)} left(n^{frac{1}{n}}-1right)^n$ is a finite number as well as $frac{e^{N(varepsilon)+1}}{1-varepsilon}$. As a result $$0<sumlimits_{ngeq1} left(n^{frac{1}{n}}-1right)^n<infty$$
  i.e. converging.

$sum_{ngeq 1}left(sqrt[n]{n}-1right)^n $ is convergent, since for any $ngeq 2$ we have
$$ sqrt[n]{n} = sqrt[n]{frac{n}{n-1}cdotfrac{n-1}{n-2}cdotldotscdotfrac{2}{1}cdotfrac{1}{1}}leq text{AM}left(1+tfrac{1}{n-1},ldots,1+tfrac{1}{2},2,1right)=1+frac{H_{n-1}}{n} $$
and by the Cauchy-Schwarz inequality $H_nleq frac{pi}{sqrt{6}}sqrt{n} $. The series $sum_{ngeq 1}frac{C}{n^{n/2}}$ is trivially convergent.