Question

Let $n$ be the smallest positive integer such that $1+\frac{1}{2}+\frac{1}{3}+\ldots+\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,
$\frac{\log _{e}(1+x)}{x} < 1$
$\therefore \frac{\log _{e}\left(1+\frac{1}{x}\right)}{1 / x}<1$
$\log _{e}\left(\frac{x+1}{x}\right) <\frac{1}{x}$
$\log _{e}(x+1)-\log _{e} x<\frac{1}{x}$
$\log _{e} 2-\log _{e} 1<1$
$\log _{e} 3-\log _{e} 2 <\frac{1}{2}$
$\log _{e}(n+1)-\log _{e} n <\frac{1}{n}$
$\log _{e}(n+1)-\log _{e} 1<1+\frac{1}{2}+\frac{1}{3}+...+\frac{1}{n}$
$\log _{e}(n+1) \leq 4$
$\left[\because 1+\frac{1}{2}+\frac{1}{3}+...+\frac{1}{n} \geq 4\right]$
$n+1< e^{4}$
$n< e^{4}-1$
$n<(2.7)^{4}-1 [\because e=2.718]$
$n<54.59-1$
$n<53.59$