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Show that $\displaystyle\sum^\infty_{n=1} \frac 1 {e^{\sqrt n}}$ converges.

My attempt:

Using limit ratio test: $\displaystyle\lim_{n\to\infty} \frac {e^{\sqrt n}} {e^{\sqrt {n+1}}}=\lim_{n\to\infty}\frac {e^{\frac n 2}} {e^{\frac {n+1} 2}}=\lim_{n\to\infty}\frac {e^{\frac n 2}} {e^{\frac n 2+\frac 1 2}}=\lim_{n\to\infty}\frac 1 {e^{\frac n 2+\frac 1 2-\frac n 2}}=e^{-0.5}$

And since $e^{-0.5}<1$ the series converges.

Is that alright?

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    $\begingroup$ In the first equality, are you saying that $$ e^{\sqrt n} = e^{n/2}? $$ If so then no, your proof is not correct. $\endgroup$ Commented Mar 3, 2015 at 21:10
  • $\begingroup$ @Omnomnomnom uh why that's not correct? $\endgroup$
    – GinKin
    Commented Mar 3, 2015 at 21:11
  • $\begingroup$ $e^{\sqrt n} \neq \sqrt{e^n}$. If we had the latter expression, we could say that $$ \sqrt{e^n} = (e^n)^{1/2} = e^{n/2} $$ $\endgroup$ Commented Mar 3, 2015 at 21:16

5 Answers 5

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The correct application of the ratio test would be as follows: $$ \begin{align*} \lim_{n\to\infty} \frac {e^{\sqrt n}} {e^{\sqrt {n+1}}}&= \lim_{n \to \infty} e^{\sqrt n - \sqrt{n+1}} \\ & = \exp\left[ \lim_{n \to \infty} \sqrt n - \sqrt{n + 1} \right] \\ & = \exp \left[ \lim_{n \to \infty} \frac{-1}{\sqrt n + \sqrt{n + 1}} \right] = 1 \end{align*} $$ So, the ratio test fails. In fact, we could use the comparison test with, for example, $\sum 1/n^2$. It suffices to note that $$ \lim_{n \to \infty} \frac{1/e^{\sqrt n}}{1/n^2} = 0 $$ To prove the above, we have $$ \lim_{n \to \infty} \frac{1/e^{\sqrt n}}{1/n^2} = \lim_{n \to \infty} \frac{n^2}{e^{\sqrt n}} \overset{u = \sqrt n}{=} \lim_{u \to \infty} \frac{(u^2)^2}{e^u} = \lim_{u \to \infty} \frac{u^4}{e^u} $$ from there, apply L'Hospital's rule.

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The condensation test works here: if $f(n)$ is positive and decreasing, then $\sum f(n)$ converges if and only if $\sum 2^n f(2^n)$ converges.

If $f(n) = e^{-\sqrt{n}}$, then $2^n f(n) = 2^n e^{-\sqrt{2^n}} = e^{n \ln 2 -2^{n/2}} $. Since $2^n > n^4$ for $n \ge 17$ (see a proof in my answer here: Prove that $n^k < 2^n$ for all large enough $n$), $ 2^{n/2} > n^2 $ so $e^{n \ln 2 -2^{n/2}} <e^{n \ln 2 -n^2} <e^{-n(n- \ln 2)} $ and the sum of these clearly converges.

Note: The result I proved there is this:

If $n$ and $k$ are integers and $k≥2$ and $n≥k^2+1$, then $2^n>n^k$.

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$$ \begin{split} \displaystyle\sum^\infty_{n=1} \frac 1 {e^{\sqrt n}}&=\ \underbrace{e^{-\sqrt1}+e^{-\sqrt2}+e^{-\sqrt3}}_{3\text{ terms}} + \underbrace{e^{-\sqrt4}+e^{-\sqrt5}+e^{-\sqrt6}+e^{-\sqrt7}+e^{-\sqrt8}}_{5\text{ terms}} + \underbrace{e^{-\sqrt9}+\cdots + e^{-\sqrt{15}}}_{7\text{ terms}} + \cdots\\ &<\ 3\cdot e^{-\sqrt1}+5\cdot e^{-\sqrt4}+7\cdot e^{-\sqrt9}+\cdots+(2k+1)\cdot e^{-\sqrt{k^2}}+\cdots\\ &=\ 3\cdot e^{-1}+5\cdot e^{-2}+7\cdot e^{-3}+\cdots+(2k+1)\cdot e^{-k}+\cdots\\ &<\ 4\cdot e^{-1}+8\cdot e^{-2}+16\cdot e^{-3}+\cdots2^{k+1}\cdot e^{-k}+\cdots\\ &=\ \frac{4\cdot e^{-1}}{1-\frac{2}{e}}<\infty. \end{split} $$

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  • $\begingroup$ Why is it equal to: $\frac{4\cdot e^{-1}}{1-\frac{2}{e}}$? $\endgroup$
    – GinKin
    Commented Mar 3, 2015 at 22:05
  • $\begingroup$ The last step uses the formula for the sum of an infinite geometric series: $a+ar+ar^2+\cdots=\frac{a}{1-r}$. The series here has first term $4e^{-1}$ and common ratio $\frac{2}{e}$ (which has absolute value less than $1$, as required by the formula). $\endgroup$
    – Steve Kass
    Commented Mar 3, 2015 at 23:06
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The step

$$\lim_{n\to\infty} \frac {e^{\sqrt n}} {e^{\sqrt {n+1}}}=\lim_{n\to\infty}\frac {e^{\frac n 2}} {e^{\frac {n+1} 2}}$$

desperately needs justification.

Since $$\frac {e^{\sqrt n}} {e^{\sqrt {n+1}}} = e^{\sqrt{n} - \sqrt{n+1}}$$ and $$\lim_{n \to \infty} \left( \sqrt{n} - \sqrt{n+1} \right) = \lim_{n \to \infty} \frac{-1}{\sqrt{n} + \sqrt{n+1}} = 0$$ you actually have $$\lim_{n \to \infty} \frac{e^{\sqrt n}} {e^{\sqrt {n+1}}} = 1.$$

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    $\begingroup$ I love the "desperately". $\endgroup$ Commented Mar 3, 2015 at 21:17
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A very simple answer just occurred to me.

For $\sum^\infty_{n=1} \frac 1 {e^{\sqrt n}} $, by looking at the first few terms of the power series, $e^{\sqrt n} \gt 1+\sqrt n +\frac{(\sqrt n)^2}{2} +\frac{(\sqrt n)^3}{6} \gt\frac{(\sqrt n)^3}{6} =\frac{n^{3/2}}{6} $ so that $\sum^N_{n=1} \frac 1 {e^{\sqrt n}} \lt \sum^N_{n=1} \frac 6 {n^{3/2}} $ which converges by the comparison test.

Note that this method can show that $\sum^N_{n=1} \frac 1 {e^{n^a}} $ converges for any $a > 0$ by going far enough out in the power series: $e^{n^a} \gt \frac{(n^a)^m}{m!} = \frac{n^{am}}{m!} $, so if we choose $m \gt \frac1{a}$, this shows that the series converges.

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  • $\begingroup$ What's stopping you from the taking the remainder like this for any series? $\endgroup$
    – shinzou
    Commented Dec 9, 2016 at 0:10
  • $\begingroup$ What remainder? $\endgroup$ Commented Dec 9, 2016 at 14:23
  • $\begingroup$ I meant the smaller part of the series, the value before the actual remainder. $\endgroup$
    – shinzou
    Commented Dec 9, 2016 at 17:45
  • $\begingroup$ Nothing. The exponential series has the nice property that all its coefficients are positive, so any particular one can be chosen. For alternating series, like for $e^{-x}, \sin(x), \cos(x)$, this does not work. $\endgroup$ Commented Dec 9, 2016 at 17:55

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