Question

Let n be the smallest positive integer such that $1+\frac{1}{2}+\frac{1}{3}+....+\frac{1}{n}\ge 4$. Which one of the following statements is true?

• A. $20<n\le 60$
• B. $60<n\le 80$
• C. $80<n\le 100$
• D. $100<n\le 120$

Answer 1

The correct option is A $20<n\le 60$

Let $S_n=\frac{1}{1}+\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+\cdots +\frac{1}{n}$

The given sequence is a Harmonic Progression and hence, its sum needs to be estimated rather than calculated.

This estimation is done replacing all numbers by their nearest lower and higher multiples of $\frac{1}{2}$ for minimum and maximum value estimation.

$1+\frac{1}{2}+\frac{1}{2}+\frac{1}{4}+\frac{1}{4}+\frac{1}{4}+\frac{1}{4}+\frac{1}{8}+\cdots >S_n>1+\frac{1}{2}+\frac{1}{4}+\frac{1}{4}+\frac{1}{8}+\frac{1}{8}+\dots$

referring to the options, we can see that the first boundary point is $n=60$.

Considering only the minimum bound as per our requirement,

$S_{60}>1+\frac{1}{2}+2\times \frac{1}{4}+4\times \frac{1}{8}+8\times \frac{1}{16}+16\times \frac{1}{32}+(60-32)\times \frac{1}{64}$

$\Rightarrow S_{60}>1+2.5+\frac{7}{16}$

$\Rightarrow S_{60}>4-\frac{1}{16}$

Now, in the minimum bound calculation, if we just remove an approximation and replace the actual value, we get

$S_{60}>4-\frac{1}{16}+\frac{1}{3}-\frac{1}{4}$

$\Rightarrow S_{60}>4-\frac{1}{16}+\frac{1}{12}$

$\Rightarrow S_{60}>4+\frac{1}{48}$

$\Rightarrow S_{60}>4$

Hence, the minimum value of $n$ for equality $\frac{1}{1}+\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+\cdots +\frac{1}{n}\ge 4$ to exist is less than $60$ and hence, the correct option is $20<n\le 60$