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.

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.

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TopNewestIt's really simple. Hint: Try grouping terms. – Paramjit Singh · 3 years, 4 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, 5 months ago

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B. – Paramjit Singh · 3 years, 5 months agoLog in to reply

– Cody Johnson · 3 years, 5 months ago

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 \(n=5\). \(\frac{1709}{1085}<\frac52\).Log in to reply

– Paramjit Singh · 3 years, 5 months ago

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

– Cody Johnson · 3 years, 4 months ago

Oops.Log in to reply