Harmonic Limit

Calculus Level 3

lim ⁡ n → ∞ n ( H n − log ⁡ n − γ ) \large \lim_{n\to\infty} n(H_{n}-\log{n}-\gamma)

Evaluate the limit above.

Notation: H n H_n denotes the n n th harmonic number .


The answer is 0.5.

Jim Chale by Jim chale , 20, Greece

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1 solution

Star Chou
Jul 10, 2018

The asymptotic approximation of the n n th harmonic number is H n = log ⁡ n + γ + 1 2 n + O ( 1 n 2 ) , H_n = \log n + \gamma + \frac{1}{2n} + O \left (\frac{1}{n^2} \right ), which can be derived from Euler's summation formula. Hence lim ⁡ n → ∞ n ( H n − log ⁡ n − γ ) = lim ⁡ n → ∞ [ 1 2 + O ( 1 n ) ] = 1 2 . \lim_{n \to \infty} n(H_n - \log n - \gamma) = \lim_{n \to \infty} \left [ \frac{1}{2} + O \left (\frac{1}{n} \right ) \right ] = \frac{1}{2}.

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