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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.

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    $\begingroup$ Do you know the Taylor expansion for $e^x$? $\endgroup$ – Clayton Sep 13 '16 at 17:21
  • $\begingroup$ @Clayton Yes, I do (but it's no obvious to me what I should do with it to solve this problem). $\endgroup$ – user347616 Sep 13 '16 at 17:22
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    $\begingroup$ Clayton suggested to take advantage from the expansion $ \sqrt[n]{e}-1=1/n+O(1/n^2)$ to infer on the behaviour of the series. The harmonic series is divergent $\endgroup$ – guestDiego Sep 13 '16 at 17:30
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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.

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

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    $\begingroup$ An easy way to see it: the limit is just the definition of $\left.\frac{de^x}{dx}\right|_{x=0} = e^0 = 1$. $\endgroup$ – Winther Sep 13 '16 at 17:24
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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.

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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 $a\le 1$.

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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.

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