Q57 / 90 · KVPY 2017 (19 Nov SB/SX) · Mathematics
Let $\mathrm{n}$ be the smallest positive integer such that $1+\frac{1}{2}+\frac{1}{3}+\ldots \ldots+\frac{1}{\mathrm{n}} \geq 4$
Which one of the following statements is true?
- Correct answerA$20 < \mathrm{n} \leq 60$
- B$60 < \mathrm{n} \leq 80$
- C$80 < \mathrm{n} \leq 100$
- D$100 < \mathrm{n} \leq 120$
Solution
$\begin{array}{l}
1+x < 1+x+\frac{x^{2}}{2 !}+\frac{x^{3}}{3 !}+\ldots \ldots \ldots \\
1+x < e^{x}
\end{array}$
$\ln (1+x) < x$
$\mathrm{x}=\frac{1}{\mathrm{y}}$
$\ln \left(1+\frac{1}{y}\right) < \frac{1}{y}$
$\ln (y+1)-\ln y < \frac{1}{y}$
put $\mathrm{y}=1$
$\ln (2)-\ln (1) < \frac{1}{1}$
put $=y=2$
$\ln 3-\ln 2 < \frac{1}{2}$
put $=y=3$
$\ln 4-\ln 3 < \frac{1}{3}$
put $y=n$
$\ln (n+1)-\ln n < \frac{1}{4}$
$\ln (n+1)-\ln (1) < 1+\frac{1}{2}+\frac{1}{3}+\ldots \frac{1}{n}$
$\ln (n+1) \leq 4$
$n \leq e^{4}-1$
$n \leq 60$
Hence, the correct answer is option: 'A'.