How to do this summation?

How to do this summation?

$$\sum_{n=1}^\infty \log_{2^\frac{n}{2^n}}256=?$$

All I'm getting is 8(2 + 2 + 8/3 + 16/4 + .... ) which is a diverging series.

-
$$\log_{2^\frac{n}{2^n}}256=\frac{2^{n+3}}{n}$$ – J. M. Jul 12 '12 at 11:32

A single term of series is $$\log_{2^{\frac{n}{2^n}}} 256=\frac{\log_2 256}{\log_2 2^{\frac{n}{2^n}}}=\frac{2^n8}{n}$$ First condition for series to be convergent is that it's terms should converge to $0$, but here $\frac{2^n8}{n}\to \infty$ as $n\to \infty \implies$ this series is not convergent.
It doesn't even have to be a base-2 logarithm, since $\log_b a=\dfrac{\log_c a}{\log_c b}$ for $c$ positive and $\neq 1$. – J. M. Jul 12 '12 at 11:37
$2^n$ grows much faster than $n$ when $n$ increases, thus $\frac{2^n}{n}$ diverges to $\infty$ (you can also apply L'Hopital's Rule) and for the base-2 part, you can choose whatever base you want(if it diverges for base-2, it will diverge for every other base.) – Aang Jul 12 '12 at 11:41
You have $$\log_{2^{n/2^n}}(256) = {{\log(256)}\over {\log(2^{n/2^n})}}= {\log(256)\over n/2^n\log(2)} = {8\cdot 2^n\over n }$$ A series with these terms will diverge, since the terms themselves will become unbounded.