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I want to prove that for $k>0$:

$ 2^k \geq \frac{-1}{\log_2(1-\frac{1}{2^k})}$

I've plotted both functions and it seems to be the case for k>0.

In fact, it would also be nice to see that:

$ \frac{-1}{\log_2(1-\frac{1}{2^k})} \geq 2^{k-1}$

Thank you!

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Are the logs $\log_2$ instead of $log2$? –  Anon Feb 1 '13 at 2:08
    
Yes, sorry, this is $\log_2$. I've corrected the question. –  user60502 Feb 1 '13 at 2:09
    
$$ \log_2 \left({ \frac{2^k - 1}{2^k} }\right) = \log_2 \left( {2^k - 1} \right) - \log_2 \left(2^k\right)$$ –  hjpotter92 Feb 1 '13 at 2:24
    
Thank you "Back in a Flash", the second term reduces to k. But now what? –  user60502 Feb 1 '13 at 2:28
    
I reached this: $\log_2(2^k-1)^{2k} \ge \log_2 2^{k2^k} -1$ –  user60502 Feb 1 '13 at 2:30
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1 Answer

With $t = 1 - 2^{-k}$ (so $0 < t < 1$ for $k > 0$), your inequality becomes $1/(1-t) \ge -1/\log_2(t)$, or $\log_2(t) \le -1+t$. In fact $\log_2(t) = \ln(t)/\ln(2)$ and $\ln(2) < 1$ so $\log_2(t) < \ln(t)$, and $\ln(t) \le -1+t$.

Your "it would be nice" is not true for $0 < k < 1$. It is true for $k \ge 1$.

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Excellent! Thank you Robert! How trivial it seems when one sees the solution :-( –  user60502 Feb 1 '13 at 2:36
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