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.
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.
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.
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.