Question

Let $n$ be the smallest positive integer such that $1+\frac{1}{2}+\frac{1}{3}+\dots +\frac{1}{n}\ge 4$ 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

Answer: (A)

We know,
$\frac{{\log }_e(1+x)}{x}<1$
$\therefore \frac{{\log }_e\left(1+\frac{1}{x}\right)}{1/x}<1$
${\log }_e\left(\frac{x+1}{x}\right)<\frac{1}{x}$
${\log }_e(x+1)-{\log }_{e}x<\frac{1}{x}$
${\log }_{e}2-{\log }_{e}1<1$
${\log }_{e}3-{\log }_{e}2<\frac{1}{2}$
${\log }_e(n+1)-{\log }_{e}n<\frac{1}{n}$
${\log }_e(n+1)-{\log }_{e}1<1+\frac{1}{2}+\frac{1}{3}+...+\frac{1}{n}$
${\log }_e(n+1)\le 4$
$\left[\because 1+\frac{1}{2}+\frac{1}{3}+...+\frac{1}{n}\ge 4\right]$
$n+1<e^4$
$n<e^4-1$
$n<(2.7{)}^4-1[\because e=2.718]$
$n<54.59-1$
$n<53.59$