Study the convergence of a series I am studying the convergence of the series $$\sum\limits_{n>=0} \left(\sqrt[n]{n} - 1\right)^n$$
I just know that the limit tend to $0$, but I don't know how to prove the convergence.
 A: $\sum_{n\geq 1}\left(\sqrt[n]{n}-1\right)^n $ is convergent, since for any $n\geq 2$ we have
$$ \sqrt[n]{n} = \sqrt[n]{\frac{n}{n-1}\cdot\frac{n-1}{n-2}\cdot\ldots\cdot\frac{2}{1}\cdot\frac{1}{1}}\leq \text{AM}\left(1+\tfrac{1}{n-1},\ldots,1+\tfrac{1}{2},2,1\right)=1+\frac{H_{n-1}}{n} $$
and by the Cauchy-Schwarz inequality $H_n\leq \frac{\pi}{\sqrt{6}}\sqrt{n} $. The series  $\sum_{n\geq 1}\frac{C}{n^{n/2}}$ is trivially convergent.
A: 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
$$n-1=
\left(n^{\frac{1}{n}}\right)^{n}-1=\\
\left(n^{\frac{1}{n}}-1\right)\left(n^{\frac{1}{n}(n-1)}+n^{\frac{1}{n}(n-2)}+...+n^{\frac{1}{n}2}+n^{\frac{1}{n}}+1\right)\geq ...$$
using AM-GM
$$... \geq\left(n^{\frac{1}{n}}-1\right)\left((n-1)\sqrt[n-1]{n^{\frac{1}{n}(n-1)}\cdot n^{\frac{1}{n}(n-2)}\cdot ...\cdot n^{\frac{1}{n}2}\cdot n^{\frac{1}{n}}}+1\right)=\\
\left(n^{\frac{1}{n}}-1\right)\left((n-1)\sqrt[n-1]{n^{\frac{1}{n}(n-1+n-2+...+1)}}+1\right)=
\\
\left(n^{\frac{1}{n}}-1\right)\left((n-1)\sqrt[n-1]{n^{\frac{1}{n}\frac{(n-1)n}{2}}}+1\right)=\\
\left(n^{\frac{1}{n}}-1\right)\left((n-1)\sqrt{n}+1\right)$$
Or
$$0< n^{\frac{1}{n}}-1\leq \frac{n-1}{(n-1)\sqrt{n}+1}<\frac{1}{\sqrt{n}} \tag{1}$$

From $(1)$ we see that $\lim\limits_{n\rightarrow\infty} \left(n^{\frac{1}{n}}-1\right)=0$, which means 


*

*$0< \left(n^{\frac{1}{n}}-1\right)^n<\frac{1}{n^{\frac{n}{2}}}$, thus $0<\sum\limits_{n\geq1} \left(n^{\frac{1}{n}}-1\right)^n \leq \sum\limits_{n\geq1} \frac{1}{n^{\frac{n}{2}}}$. The latter is converging by ratio test $$\lim\limits_{n\rightarrow\infty}\frac{\frac{1}{(n+1)^{\frac{n+1}{2}}}}{\frac{1}{n^{\frac{n}{2}}}}=
\lim\limits_{n\rightarrow\infty}\frac{n^{\frac{n}{2}}}{(n+1)^{\frac{n+1}{2}}}=
\lim\limits_{n\rightarrow\infty}\frac{1}{\left(1+\frac{1}{n}\right)^{\frac{n}{2}}\cdot \sqrt{n+1}}=\frac{1}{\sqrt{e}\cdot \lim\limits_{n\rightarrow\infty}\sqrt{n+1}}=0$$

*or simpler version from the definition of the limit, for $\forall n> N(\varepsilon) \Rightarrow 0<\left(n^{\frac{1}{n}}-1\right)<\varepsilon <1$, thus $0<\left(n^{\frac{1}{n}}-1\right)^n<\varepsilon^n$ and $$0<\sum\limits_{n\geq1} \left(n^{\frac{1}{n}}-1\right)^n <
\sum\limits_{n=1}^{N(\varepsilon)} \left(n^{\frac{1}{n}}-1\right)^n + \sum\limits_{n=N(\varepsilon)+1} \varepsilon^n=
\sum\limits_{n=1}^{N(\varepsilon)} \left(n^{\frac{1}{n}}-1\right)^n + \frac{e^{N(\varepsilon)+1}}{1-\varepsilon}$$
we just need to find the very first $N(\varepsilon)$ for $0<\varepsilon<1$ and $\sum\limits_{n=1}^{N(\varepsilon)} \left(n^{\frac{1}{n}}-1\right)^n$ is a finite number as well as $\frac{e^{N(\varepsilon)+1}}{1-\varepsilon}$. As a result $$0<\sum\limits_{n\geq1} \left(n^{\frac{1}{n}}-1\right)^n<\infty$$
i.e. converging.

