Convergence of the series $\sqrt[n]n-1$ Let $a_n=\sqrt[n]n-1$. Does $\sum_{n=1}^\infty a_n$ converge? 
 A: $$\sqrt[n]{n}-1=\frac{n-1}{\sqrt[n]{n^{n-1}}+\sqrt[n]{n^{n-2}}+...+1}\geq \frac{n-1}{n\sqrt[n]{n^{n-1}}}=\frac{n-1}{n}\frac{1}{n^{\frac{n-1}{n}}}=\frac{n-1}{n}\frac{\sqrt[n]{n}}{n}>\frac{n-1}{n}\frac{1}{n}$$
Now use that for all $n >2$ we have
$$\frac{n-1}{n}\frac{1}{n}>\frac{1}{2}\frac{1}{n} \,,$$
or limit compare it to the harmonic series.
A: For $n\ge2$,
$$
\begin{align}
n(\sqrt[n]{n}-1)
&\ge n(2^{1/n}-1)\\
&\to\log(2)
\end{align}
$$
Thus, for some $N$, if $n\ge N$, then $n(\sqrt[n]{n}-1)\ge\frac12$. Therefore, for $n\ge N$,
$$
\sqrt[n]{n}-1\ge\frac1{2n}
$$
Thus, the series diverges by comparison to the harmonic series.
A: Hint: $$\sqrt[n]n  = e^{\frac{\log n}{n}} > 1+\frac{\log n}{n}$$
A: $$a_n=n^{1/n}-1= e^{\log{n}/n}-1 \sim \frac{\log{n}}{n}$$
The sum diverges by comparison with the sum of $b_n=1/n$.
A: From $\left(1+\frac1n\right)^n\to e$, we conclude 
$$ \left(1+\frac1n\right)^n<3<n=(1+a_n)^n$$
for almost all $n$. Hence $a_n>\frac1n$ for almost all $n$ and $\sum a_n$ diverges.
Remark: We don't even need the introduction of $e$. The observation
$$\left(1+\frac1n\right)^n=\sum_{k=0}^n{n\choose k}n^{-k}\le \sum_{k=0}^n\frac1{k!}<1+\sum_{k=1}^ \infty 2^{1-k} =3$$
suffices for an "elementary" approach.
A: Note that for any $n\ge3$:
$$ n>e$$
and so,
$$ n^{1/n} > e^{1/n}$$
And since we know:
$$ e^x = \sum_{k=0}^\infty \frac{x^k}{k!}$$ we can say:
$$ e^{1/n} = \sum_{k=0}^\infty \frac{1}{k!\;n^k} = 1 + \frac{1}{n} + \frac{1}{2n^2}+\cdots$$
Which, for $n\in\Bbb N$, is surely larger than the truncated series $1+1/n$, and so follows the strict inequality:
$$n^{1/n} > e^{1/n} > 1 + \frac{1}{n}$$
Subtract one from both sides, yielding:
$$\sqrt[n]{n}-1 = n^{1/n} - 1>\frac1n$$
Which is true $\forall n\ge 3$. Evaluate the sum:
$$ \sum_{n\ge 3} (\sqrt[n]{n}-1) > \sum_{n\ge 3} \frac{1}{n}$$ 
Thus, $\sum a_n$ is larger than the divergent harmonic sum, and is divergent too.
