Some Harmonic sum

Calculus Level 5

∑ n = 1 ∞ H n 2 n + 1 ( n + 1 ) = ? \displaystyle \large \sum_{n=1}^{\infty}\dfrac{H_{n}}{2^{n+1} (n+1)} = ?

Notation : H n H_n denotes the n th n^\text{th} harmonic number , H n = 1 + 1 2 + 1 3 + ⋯ + 1 n H_n = 1 + \dfrac12 + \dfrac13 + \cdots + \dfrac1n .


The answer is 0.2402.

Harsh Shrivastava by Harsh Shrivastava , 20, India

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1 solution

Chew-Seong Cheong
Jun 13, 2016

The generating function of harmonic numbers is as below:

∑ n = 1 ∞ H n x n = − ln ⁡ ( 1 − x ) 1 − x for − 1 ≤ x < 1 ∫ 0 1 2 ∑ n = 1 ∞ H n x n d x = ∫ 0 1 2 − ln ⁡ ( 1 − x ) 1 − x d x RHS by integration by parts ∑ n = 1 ∞ H n 2 n + 1 ( n + 1 ) = ln ⁡ 2 ( 1 − x ) ∣ 0 1 2 − ∫ 0 1 2 − ln ⁡ ( 1 − x ) 1 − x d x = 1 2 ⋅ ln ⁡ 2 ( 1 − x ) ∣ 0 1 2 = ln ⁡ 2 2 2 ≈ 0.2402 \begin{aligned} \sum_{n=1}^\infty H_n x^n & = \frac{- \ln (1-x)}{1-x} \quad \text{for} -1 \le x < 1 \\ \int_0^\frac 12 \sum_{n=1}^\infty H_n x^n \ dx & = \color{#3D99F6}{ \int_0^\frac 12 \frac{- \ln (1-x)}{1-x} dx} \quad \quad \small \color{#3D99F6}{\text{RHS by integration by parts}} \\ \sum_{n=1}^\infty \frac{H_n}{2^{n+1}(n+1)} & = \ln^2 (1-x) \bigg|_0^\frac 12 - \int_0^\frac 12 \frac{- \ln (1-x)}{1-x} dx \\ & = \frac 12 \cdot \ln^2 (1-x) \bigg|_0^\frac 12 \\ & = \frac {\ln^2 2}2 \approx \boxed{0.2402} \end{aligned}

5 Helpful 0 Interesting 1 Brilliant 0 Confused

IBP is not necessary, since − 1 1 − x ln ⁡ ( 1 − x ) = f ( x ) f ′ ( x ) -\frac{1}{1-x}\ln(1-x) = f(x)f'(x) is the derivative of 1 2 f ( x ) 2 = 1 2 ln ⁡ 2 ( 1 − x ) \tfrac12f(x)^2 = \tfrac12\ln^2(1-x) .

Mark Hennings - 5 years ago

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Thanks a lot.

Chew-Seong Cheong - 5 years ago

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