Show that this series converges 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
 A: $\log(2^n) = n \log (2)$, so we consider $\displaystyle\sum\dfrac{\sqrt{n}}{2^{n/2}}$. But any polynomial over an exponential converges, right?
A: $\log(2^n) = n \log(2)$.  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.
A: 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.
A: The ratio test or root test will do this very nicely.  
