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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.
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.
Since \begin{align*} \lim_{n\rightarrow\infty}\frac{\sqrt[n]{e}-1}{\frac{1}{n}} =1, \end{align*} the series is divergent.
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.
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 $a\le 1$.
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.
