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# Prove this

$\large\dfrac1{2^m} \sum_{k=1}^{2^m} \dfrac{f(k)} k > \dfrac23.$

Let $$f(n)$$ be the greatest odd divisor of $$n$$. Prove that for all positive integers $$m$$, the inequality above holds true.

Note by Yash Saxena
2 years, 1 month ago

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Induct on $$m$$; this is obvious for $$m = 1$$. Assume this is true for $$m = M$$, and we will prove this for $$m = M+1$$.

Note that $$f(k) = f(2k)$$ and $$f(2k-1) = 2k-1$$. Thus,

\begin{align*} \frac{1}{2^{M+1}} \sum_{k=1}^{2^{M+1}} \frac{f(k)}{k} &= \frac{1}{2^{M+1}} \sum_{k=1}^{2^M} \left( \frac{f(2k-1)}{2k-1} + \frac{f(2k)}{2k} \right) \\ &= \frac{1}{2^{M+1}} \sum_{k=1}^{2^M} \left( \frac{2k-1}{2k-1} + \frac{f(k)}{2k} \right) \\ &= \frac{1}{2^{M+1}} \sum_{k=1}^{2^M} \left( 1 + \frac{1}{2} \cdot \frac{f(k)}{k} \right) \\ &= \frac{1}{2^{M+1}} \sum_{k=1}^{2^M} 1 + \frac{1}{2^{M+1}} \sum_{k=1}^{2^M} \frac{1}{2} \cdot \frac{f(k)}{k} \\ &= \frac{1}{2^{M+1}} \cdot 2^M + \frac{1}{2^{M+2}} \sum_{k=1}^{2^M} \frac{f(k)}{k} \\ &> \frac{1}{2} + \frac{1}{4} \cdot \frac{2}{3} \\ &= \frac{2}{3} \end{align*}

In the first equality, we unroll the summation so we're summing two terms at a time. The second equality uses the aforementioned facts $$f(k) = f(2k)$$ and $$f(2k-1) = 2k-1$$. The rest should be fairly straightforward; the inequality invokes the inductive hypothesis.

This completes the proof for $$m = M+1$$, and thus finishes the proof by induction.

- 2 years, 1 month ago

Was writing up a proof, but your method is surely more elegant. Well done!

- 2 years, 1 month ago