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.

#Range #Goldbach'sConjurersGroup

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`2^{34}`	$2^{34}$
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`\frac{2}{3}`	$\frac{2}{3}$
`\sqrt{2}`	$\sqrt{2}$
`\sum_{i=1}^3`	$\sum_{i=1}^3$
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Comments

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 - 7 years, 3 months ago

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

A Brilliant Member - 7 years, 3 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 $n=5$ . $\frac{1709}{1085}<\frac52$ .

Cody Johnson - 7 years, 3 months ago

@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 $\displaystyle \sum_{i=1}^{2^n-1} \frac{1}{i}$ .

A Brilliant Member - 7 years, 3 months ago

@A Brilliant Member – Oops.

Cody Johnson - 7 years, 3 months ago

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

A Brilliant Member - 7 years, 3 months ago

Math	Appears as
Remember to wrap math in `\(` ... `\)` or `\[` ... `\]` to ensure proper formatting.
`2 \times 3`	$2 \times 3$
`2^{34}`	$2^{34}$
`a_{i-1}`	$a_{i-1}$
`\frac{2}{3}`	$\frac{2}{3}$
`\sqrt{2}`	$\sqrt{2}$
`\sum_{i=1}^3`	$\sum_{i=1}^3$
`\sin \theta`	$\sin \theta$
`\boxed{123}`	$\boxed{123}$