HMMT November

Algebra Difficulty 4.9 AIME Find the answer

For the specific example $M=5$ , find a value of $k$ , not necessarily the smallest, such that $\sum_{n=1}^{k} \frac{1}{n}>M$ . Justify your answer.

Your answer

A number or a short expression. Fractions can be typed as 3/2, and spacing doesn't matter.

Solution

Note that $\frac{1}{n+1}+\frac{1}{n+2}+\ldots+\frac{1}{2n}>\frac{1}{2n}+\ldots+\frac{1}{2n}=\frac{1}{2}$ . Therefore, if we apply this to $n=1,2,4,8,16,32,64,128$ , we get $\left(\frac{1}{2}\right)+\left(\frac{1}{3}+\frac{1}{4}\right)+\left(\frac{1}{5}+\frac{1}{6}+\frac{1}{7}+\frac{1}{8}\right)+\ldots+\left(\frac{1}{129}+\ldots+\frac{1}{256}\right)>\frac{1}{2}+\ldots+\frac{1}{2}=4$ so, adding in $\frac{1}{1}$ , we get $\sum_{n=1}^{256} \frac{1}{n}>5$ so $k=256$ will suffice.

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