2
$\begingroup$

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.

  • 3
    $\log_{2^\frac{n}{2^n}}256=\frac{2^{n+3}}{n}$2012-07-12

2 Answers 2

4

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.

  • 1
    $2^n$ 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.)2012-07-12
2

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.