Convergence of $\sum_{n=1}^{\infty}(\sqrt[n]{e}-1)$ How would one prove that the series 
$$\sum_{n=1}^{\infty}(\sqrt[n]{e}-1)$$ 
diverges? The Root and Ratio test are useless here. It is also not so obvious to me how I could bound the given series by a smaller series which diverges. 
A friend a mine told me that I can use the Cauchy Condensation Test and I got to this point 
$$\sum_{n=1}^{\infty}(\sqrt[n]{e}-1) \sim \sum_{n=1}^{\infty}2^n(\sqrt[2n]{e}-1)$$
but it is unclear to me how to continue from here. 
 A: The sequence given by
$$ a_n=\left(1+\frac{1}{n}\right)^n $$
is an increasing sequence due to the AM-GM inequality and it converges to $e$. 
It follows that
$$ \sqrt[n]{e} > 1+\frac{1}{n} $$
hence the given series is divergent by comparison with the harmonic series.
A: More generally,
if $a > 1$,
$\sum_{n=1}^{\infty}(\sqrt[n]{a}-1)
$
diverges.
Since
$e^x \ge 1+x$,
$\sqrt[n]{a}
=e^{\ln(a)/n}
\ge 1+\ln(a)/n
$,
so that
$\sqrt[n]{a}-1
\ge \ln(a)/n
$
and the sum of these
diverges.
A: Hint: $$\sqrt[n]{e}=e^{1/n}=1+\frac{1}{n}+\cdots>1+\frac{1}{n}.$$ Thus, we have $\sqrt[n]{e}-1>\frac{1}{n}$, and now we can look at partial sums to conclude the series is divergent.
A: Since
\begin{align*}
\lim_{n\rightarrow\infty}\frac{\sqrt[n]{e}-1}{\frac{1}{n}} =1,
\end{align*}
the series is divergent.
A: In THIS ANSWER, I showed using only the limit definition and Bernoulli's Inequality that the exponential function satisfies the inequalities

$$1+x\le e^x\le \frac{1}{1-x} \tag 1$$

where the left-hand side inequality of $(1)$ holds everywhere whereas the right-hand side inequality holds for $x<1$.  Therefore, we find that for $n>1$ and any $a>0$  
$$\frac{1}{n^a}\le e^{1/n^a}-1\le \frac{1}{n^a-1} \tag 2$$
Hence, we conclude that the series $\sum_{n=1}^\infty \left(e^{1/n^a}-1\right)$ converges for $a>1$ and diverges for $a\le 1$. 
