MCQ
Let $n$ be the smallest positive integer such that $1+\frac{1}{2}+\frac{1}{3}+\ldots+\frac{1}{n} \geq 4$. Which one of the following statements is true?
- ✓$20 < n \leq 60$
- B$60 < n \leq 80$
- C$80 < n \leq 100$
- D$100 < n \leq 120$
We know,
$\frac{\log _e(1+x)}{x} < 1$
$\frac{\log _e\left(1+\frac{1}{x}\right)}{1 / x}-1$
$\log _e\left(\frac{x+1}{x}\right) < \frac{1}{x}$
$\log _e(x+1)-\log _e x < \frac{1}{x}$
$\log _e 2-\log _e 1 < 1$
$\log _e 3-\log _e 2 < \frac{1}{2}$
$\log _e(n+1)-\log _e n < \frac{1}{n}$
$\log _e(n+1)-\log _e 1<1+\frac{1}{2}+\frac{1}{3}+\ldots+\frac{1}{n}$
$\log _e(n+1) \leq 4$
${\left[\because 1+\frac{1}{2}+\frac{1}{3}+\ldots+\frac{1}{n} \geq 4\right] }$
$n+1 < e^4$
$n < e^4-1$
$n < (2.7)^4-1$
$n < 5459-1$
$n < 53.59$
$\quad[\because e=2718]$
$\therefore$ Hence, option $(a)$ is correct.
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