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In order to find if the series is convergente or divergent:

$$\sum_{n=1}^{\infty}\frac{\sqrt n}{2^n}$$

I did the ratio test: $$\lim_{n\to \infty}\left|\frac{a_{n+1}}{a_n}\right|$$

I did:

$$\lim_{n\to \infty}\left|\frac{\frac{\sqrt{n+1}}{2^{n+1}}}{\frac{\sqrt{n}}{2^n}}\right|= 2 > 1 $$therefore by the ratio test it should diverge, but this series converge as wolfram alpha says

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    $\begingroup$ I'm afraid you calculated the ratio incorrectly, it should be $\frac{1}{2}$ in the limit. $\endgroup$ – vadim123 Mar 11 '15 at 22:09
  • $\begingroup$ Please do not use titles consisting only of math expressions; these are discouraged for technical reasons -- see meta. $\endgroup$ – AlexR Mar 11 '15 at 22:12
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    $\begingroup$ There was a mistake by computing the Limit. $\endgroup$ – kryomaxim Mar 11 '15 at 22:15
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    $\begingroup$ This is a polylogarithm. $\endgroup$ – Lucian Mar 11 '15 at 22:20
  • $\begingroup$ For information : $\sum_{n=1}^{\infty}\frac{\sqrt n}{2^n}=Li_{-1/2}(1/2)=1.34725375...$ where $Li$ is the function polylogarithm. $\endgroup$ – JJacquelin May 10 '15 at 21:52
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Observe that $$\lim_{n\to \infty}\left|\frac{a_{n+1}}{a_n}\right|=\lim_{n\to \infty}\left|\frac{\frac{\sqrt{n+1}}{2^{n+1}}}{\frac{\sqrt{n}}{2^n}}\right|= \lim_{n\to \infty}\left|\frac{\sqrt{n+1}}{2^{n+1}}\times\frac{2^n}{\sqrt{n}}\right|=\lim_{n\to \infty}\left|\frac{2^n}{2^{n+1}}\right|\times\lim_{n\to \infty}\left|\frac{\sqrt{n+1}}{\sqrt{n}}\right|=\frac12$$

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As noted in the comments, this series converges by the ratio test. Here we have $a_n=\dfrac{\sqrt n}{2^n}$ so that $$ \lim_{n\to\infty}\left\lvert\frac{a_{n+1}}{a_n}\right\rvert = \lim_{n\to\infty}\left\lvert\frac{\frac{\sqrt{n+1}}{2^{n+1}}}{\frac{\sqrt n}{2^n}}\right\rvert =\lim_{n\to\infty}\frac{2^n}{2^{n+1}}\frac{\sqrt{n+1}}{\sqrt n} =\frac{1}{2}\lim_{n\to\infty}\sqrt{\frac{n+1}{n}} =\frac{1}{2}\cdot 1 =\frac{1}{2} $$ It's very likely you made a typo while carrying out your original computation!

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You did the limit wrong: $$\lim_{n\to \infty}\left|\frac{\frac{\sqrt{n+1}}{2^{n+1}}}{\frac{\sqrt{n}}{2^n}}\right|= \lim\left|\frac{\frac{\sqrt{n+1}}{2^{n+1}}}{\frac{2\sqrt{n}}{2^{n+1}}}\right|=\lim \left|\frac{\sqrt{n+1}}{2\sqrt n}\right|\to \frac{1}{2} < 1$$

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    $\begingroup$ I don't think that $\frac12 > 1$^^ $\endgroup$ – AlexR Mar 11 '15 at 22:27
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You can actually show convergence by a different method (which also incidentally includes the ratio test). Note that $\sqrt(n)$ < $n$, so $\sum \sqrt(n)/(2^n)$ < $\sum n/(2^n)$ from n = 1 to $\infty$. The latter sum converges due to the ratio test criterion, so the original sum must converge.

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