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Find the ranges,ranger!

For any \(n \geq 5\), the value of \(1+\frac{1}{2} + \frac{1}{3} + \cdots + \frac{1}{2^n-1}\) lies between:

  • \(0\) and \(\frac{n}{2}\)

  • \(\frac{n}{2}\) and \(n\)

  • \(n\) and \(2n\)

  • none of the above.

Note by Paramjit Singh
3 years, 9 months ago

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It's really simple. Hint: Try grouping terms.

Paramjit Singh - 3 years, 9 months ago

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Let's find a lower bound and an upper bound, just for fun.

Lower:

\[\sum_{i=1}^n\frac1{2^i-1}\ge\sum_{i=1}^n\frac1{2^i}=1-\frac{1}{2^n}\]

Upper:

\[\frac1{2^n-1}+\sum_{i=2}^{n-1}\frac1{2^i-1}\le\sum_{i=2}^n\frac1{2^i-2}=\frac12+\frac12\sum_{i=2}^{n-1}\frac1{2^i-1}\]

Hence,

\[\sum_{i=1}^n\frac1{2^i-1}\le2-\frac1{2^n-1}\]

In total, the bound is

\[1<1-\frac{1}{2^n}\le\sum_{i=1}^n\le2-\frac1{2^n-1}<2\]

Cody Johnson - 3 years, 9 months ago

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None of the above. :( But I have the answer B.

Paramjit Singh - 3 years, 9 months ago

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Since \(n<5\), we have \(\frac{n}2>2\), so the answer is (A). Also, (B) is not the answer because consider the case of \(n=5\). \(\frac{1709}{1085}<\frac52\).

Cody Johnson - 3 years, 9 months ago

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@Cody Johnson No my friend, you treated it as \(\displaystyle \sum_{i=1}^n \frac{1}{2^i-1}\), but the question asks to bound \(\displaystyle \sum_{i=1}^{2^n-1} \frac{1}{i}\).

Paramjit Singh - 3 years, 9 months ago

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@Paramjit Singh Oops.

Cody Johnson - 3 years, 9 months ago

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