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(A) The harmonic series $H_n = 1 + \frac{1}{2} + \frac{1}{3} + \dots + \frac{1}{n}$ can be approximated by $\ln(n) + \gamma$,where $\gamma \approx 0.5

Vedclass · Accepted Answer

$20 < n \leq 60$

Vedclass · Answer

(A) The harmonic series $H_n = 1 + \frac{1}{2} + \frac{1}{3} + \dots + \frac{1}{n}$ can be approximated by $\ln(n) + \gamma$,where $\gamma \approx 0.577$ is the Euler-Mascheroni constant.We want $H_n \geq 4$,so $\ln(n) + 0.577 \approx 4$,which gives $\ln(n) \approx 3.423$.Calculating $n \approx e^{3.423} \approx 30.66$.However,using the inequality $\ln(n+1) For $H_n \geq 4$,we have $1 + \ln(n) > 4$,so $\ln(n) > 3$,which means $n > e^3 \approx 20.08$.More precisely,the value of $n$ for which the harmonic sum reaches $4$ is $n = 31$.Since $31$ falls in the range $20 < n \leq 60$,option $A$ is correct.

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?

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