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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
4 years, 1 month ago

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

- 4 years ago

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

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

- 4 years ago

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

- 4 years 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 $$n=5$$. $$\frac{1709}{1085}<\frac52$$.

- 4 years ago

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

- 4 years ago

Oops.

- 4 years ago