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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
1 year, 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. Ivan Koswara · 1 year, 1 month ago

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@Ivan Koswara Was writing up a proof, but your method is surely more elegant. Well done! Jake Lai · 1 year, 1 month ago

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