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I want to show that the following series converges.

$$\sum_{n=1}^\infty 2^{-n/2}\sqrt{\log{(2^n)}}$$

I tried to bound it from above by a convergent sequence without success. It would be appreciated if someone could help me. Thanks four your help.

hulik

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As an aside, it converges to $\sqrt{\log(2)}\mathrm{Li}_{\frac{-1}{2}}\left(\frac{1}{\sqrt{2}}\right)$, where $\mathrm{Li}$ is the polylogarithm. –  deoxygerbe Mar 5 '12 at 17:15

4 Answers 4

$\log(2^n) = n \log (2)$, so we consider $\displaystyle\sum\dfrac{\sqrt{n}}{2^{n/2}}$. But any polynomial over an exponential converges, right?

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How do you show $\sum \frac{n^k}{l^n}$ for every $k,l> 0$? –  user20869 Mar 5 '12 at 16:43
    
I think we also need $l>1$. Then the ratio test works. –  Patrick Mar 5 '12 at 21:06
    
@hulik: I only just now checked. The ratio test works, as Patrick mentioned. It is also clearly false if $l < 1$, as then the limit of the terms isn't even $0$. –  mixedmath Mar 5 '12 at 22:12

Since you mentioned the comparison test... First notice that $$2^{-n/2}\sqrt{\log{(2^n)}} = \frac{\sqrt{n\log{(2)}}}{\sqrt{2^n}} = \sqrt{\frac{n\log(2)}{2^n}},$$ so $$\sum_{n=1}^\infty 2^{-n/2}\sqrt{\log{(2^n)}} = \sqrt{\log(2)}\sum_{n=1}^\infty \sqrt{\frac{n}{2^n}}.$$ Next, notice that $2^{n/2}\geq n$ for any $n\geq 4$. Hence: $$\sum_{n=4}^\infty 2^{-n/2}\sqrt{\log{(2^n)}} = \sqrt{\log(2)}\sum_{n=4}^\infty \sqrt{\frac{n}{2^n}} \leq \sqrt{\log(2)}\sum_{n=4}^\infty \sqrt{\frac{2^{n/2}}{2^{n}}} = \sqrt{\log(2)}\sum_{n=4}^\infty \sqrt{\frac{1}{2^{n/2}}}=\sqrt{\log(2)}\sum_{n=4}^\infty \frac{1}{2^{n/4}},$$ and the last series is clearly convergent! Thus, the original series is convergent as well.

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$\log(2^n) = n \log(2)$. So, in your series, the $\sqrt{\log 2}$ is a constant that can be pulled out, i.e., it does not affect convergence or divergence. What you're left with is $$\sum_{n = 1}^\infty \frac{\sqrt{n}}{2^{n/2}}$$ Does that seem easier? Use the ratio test.

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The ratio test or root test will do this very nicely.

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