# Test for convergence $\sum_{n=1}^{\infty} \frac{1}{2^\sqrt{n}}$ [duplicate]

Possible Duplicate:
convergence of a series involving $x^\sqrt{n}$

Test for convergence $$\sum_{n=1}^{\infty} \frac{1}{2^\sqrt{n}}$$ My first thought was to use the ratio test but it's inconclusive since it yields $1$.
Are there some easy means to test the sum for convergence? Thanks!

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## marked as duplicate by GEdgar, Did, user17762, Argon, Davide GiraudoJan 12 at 22:16

This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.

## 6 Answers

Hint: For $n>2^{10}$, we have that $$\sqrt{n}\geq 2\log_2(n),$$ and so for $n>2^{10}$ $$2^{\sqrt{n}}\geq n^2.$$ Now try using the comparison test.

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 This is definitely a very good option. – Chris's wise sister Jan 12 at 21:59

A quick answer can be given by comparison with the series $\sum_{n\geq 1}1/n^2$.

Check that $\lim_{n\rightarrow +\infty} n^2 /2^\sqrt{n}=0$.

Deduce that there is a positive constant $K$ such that $\frac{1}{2^\sqrt{n}}\leq\frac{K}{n^2}$ for all $n\geq 1$.

Conclude by comparison.

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$$2^{-\sqrt n}\leqslant\int_{n-1}^n\exp(-\sqrt x\log 2).$$ The integral $\int_1^{+\infty}\exp(-\sqrt x\log 2)dx$ is convergent, using the substitution $s=\sqrt x$. We conclude by integral test, after having checked we can use it.

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This can be handled with a higher order ratio test, Raabe's test. Let $a_n = 1/2^{\sqrt{n}}$. Then $$\begin{eqnarray*} n\left(\left|\frac{a_{n+1}}{a_n}\right| - 1\right) &=& n\left(2^{\sqrt{n}-\sqrt{n+1}} - 1\right) \\ &<& n\left(2^{-1/(2\sqrt{n+1})}-1\right) \\ &<& -\frac{n}{4\sqrt{n+1}}\log 2. \end{eqnarray*}$$ In the second to last line we use the square root inequality $\sqrt{n}-\sqrt{n+1} < -1/(2\sqrt{n+1})$ for $n\ge 0$. In the last line we use the exponential inequality $e^{-x} < 1-x/2$ for $0<x<2+W(-2/e^2) = 1.59362\cdots$. Now we need only show that $$\lim_{n\to\infty} -\frac{n}{4\sqrt{n+1}}\log 2 < -1.$$

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 Raabe's test might be an option. (+1) Thanks oen! – Chris's wise sister Jan 12 at 21:56 @Chris'ssister: Glad to help. – oen Jan 12 at 21:58

So, you could use the comparison test, that one would yield results.

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Update after coment of did

The index the translation $k+1=n$, in the serie $\displaystyle\sum_{n=1}^{\infty}\frac{1}{2^{\sqrt{n}}}$ give us $$\sum_{n=1}^{\infty}\frac{1}{2^{\sqrt{n}}}= \sum_{k=0}^{\infty}\frac{1}{2^{\sqrt{k+1}}}.$$ We have by inequality $(1+x)^r\geq 2^{-r\cdot x}$ for $x>0$ large and $r>0$ that $(1+k)^{\frac{1}{2}}\geq 2^{-\frac{1}{2}k}$ implies $2^{(1+k)^{\frac{1}{2}}} \geq 2^{2^{-\frac{1}{2}k}}=2^{\frac{1}{\sqrt{2}}\cdot k}$ and $$\frac{1}{2^{(1+k)^{\frac{1}{2}}}}\leq \frac{1}{2^{\frac{1}{\sqrt{2}}\cdot k}} = \left[\frac{1}{2^{\frac{1}{\sqrt{2}}}}\right]^{k}=\left[ \frac{1}{1.63240609...}\right]^k$$ Then $$\sum_{k=0}^{\infty}\frac{1}{2^{\sqrt{1+k}}}< \sum_{k=0}^{\infty}\left[\frac{1}{2^{\frac{1}{\sqrt{2}}}}\right]^{k}<\infty$$ By comparation test the serie $\displaystyle\sum_{n=1}^{\infty}\frac{1}{2^{\sqrt{n}}}$ converge.

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We have that $(1+k)^{\frac{1}{2}}\geq 1+\frac{1}{2}\cdot k$... Really? – Did Jan 12 at 21:15
@Elias: Try $k=3$ ... – oen Jan 12 at 21:32
@did I correct may mistake. I belive now is all right. – Elias Jan 12 at 21:51
@did I correct may mistake. – Elias Jan 12 at 21:52
Still wrong. Please check what you write on simple cases. – Did Jan 12 at 22:01
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