Let n be smallest positive integer such that 1+frac{1}{2}+frac{1}{3}+ldots+frac{1}{n} ≥ 4 . Which

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Basic of Logarithms

normal

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?

$20 < n \leq 60$

$60 < n \leq 80$

$80 < n \leq 100$

$100 < n \leq 120$

(KVPY-2017)

Solution

(a)

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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