A partial harmonic series...

Calculus Level 2

lim ⁡ n → ∞ ∑ i = n A ⋅ n 1 i = ? \large \displaystyle \lim_{n\to\infty} \sum\limits_{i=n}^{A·n} \dfrac{1}{i} = ?

A A A^A A e A^e ln ⁡ ( A n ) \ln\left( \frac{A}{n}\right) ln ⁡ ( A ⋅ n ) \ln(A·n) e A e^A ln ⁡ ( n ) \ln(n) ln ⁡ ( A ) \ln(A) e A n e^ \frac{A}{n}

Tj Evert by TJ Evert , 53, USA

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1 solution

Relevant wiki: Harmonic Number

L = lim ⁡ n → ∞ ∑ k = n A n 1 k = lim ⁡ n → ∞ ( ∑ k = 1 A n 1 k − ∑ k = 1 n 1 k ) Since the n th harmonic number H n = ∑ k = 1 n 1 k = lim ⁡ n → ∞ ( H A n − H n ) and lim ⁡ n → ∞ ( H n − ln ⁡ n ) = γ = lim ⁡ n → ∞ ( ln ⁡ ( A n ) − ln ⁡ n ) where γ is the Euler-Mascheroni constant. = lim ⁡ n → ∞ ln ⁡ ( A n n ) = ln ⁡ A \begin{aligned} L & = \lim_{n \to \infty} \sum_{k=n}{An} \frac 1k \\ & = \lim_{n \to \infty} \left(\sum_{k=1}^{An} \frac 1k - \sum_{k=1}^n \frac 1k \right) & \small \color{#3D99F6} \text{Since the }n\text{th harmonic number }H_n = \sum_{k=1}^n \frac 1k \\ & = \lim_{n \to \infty} \left(H_{An} - H_n\right) & \small \color{#3D99F6} \text{and } \lim_{n \to \infty} \left(H_n - \ln n\right) = \gamma \\ & = \lim_{n \to \infty} \left(\ln (An) - \ln n \right) & \small \color{#3D99F6} \text{where } \gamma \text{ is the Euler-Mascheroni constant.} \\ & = \lim_{n \to \infty} \ln \left(\frac {An}n\right) \\ & = \boxed {\ln A} \end{aligned}

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