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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.

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$$\log_{2^\frac{n}{2^n}}256=\frac{2^{n+3}}{n}$$ – J. M. Jul 12 '12 at 11:32
up vote 4 down vote accepted

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.

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So, am I correct? This summation is not convergent, thus no finite answer? – Hyperbola Jul 12 '12 at 11:35
you got it right. :) – Aang Jul 12 '12 at 11:36
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
When n tends to infinity, (8)(2^n)/n is indeterminant, isn't it? Its of the form inf/inf – Hyperbola Jul 12 '12 at 11:39
$2^n$ 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.

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