Nice Intermediate

Calculus Level 3

lim n 1 ln n k = 1 n 1 k = ? \large \lim_{n \to \infty}\frac{1}{\ln n} \sum_{k=1}^{n} \frac{1}{k} = ?


The answer is 1.

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1 solution

Joseph Newton
Jan 22, 2018

The sum is the harmonic series, which can be approximated by: k = 1 n 1 k = 1 + 1 2 + 1 3 + + 1 n = ln n + γ \sum_{k=1}^{n} \frac{1}{k}=1+\frac{1}{2}+\frac{1}{3}+\dots+\frac{1}{n}=\ln n+\gamma where γ \gamma is the Euler–Mascheroni constant. γ 0.577 \gamma\approx0.577\dots

Therefore, we can rewrite the question: lim n 1 ln n k = 1 n 1 k = lim n 1 ln n ( ln n + γ ) = lim n ( 1 + γ ln n ) = 1 \begin{aligned}\lim_{n\to\infty}\frac{1}{\ln n}\sum_{k=1}^{n}\frac{1}{k}&=\lim_{n\to\infty}\frac{1}{\ln n}\left(\ln n+\gamma\right)\\ &=\lim_{n\to\infty}\left(1+\frac{\gamma}{\ln n}\right)\\ &=1\end{aligned} As n n approaches infinity, ln n \ln n also approaches infinity, so the γ ln n \frac{\gamma}{\ln n} approaches zero and disappears, leaving an answer of 1 \boxed{1}

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