Inspired by Patrick Corn

Calculus Level 5

∑ n = 1 ∞ ( − 1 ) n − 1 H n n \large \displaystyle\sum \limits^{\infty }_{n=1}\dfrac{( -1) ^{n-1}H_{n}}{n}

If the series above is equal to ζ ( A ) B − ( ln ⁡ D ) C E , \dfrac{\zeta ( A) }{B} -\dfrac{(\ln D)^C }{E} , where A A , B B , C C , D D and E E are all positive integers, find A + B + C + D + E A+B+C+D+E .

Notations :

  • 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 .

  • ζ ( ⋅ ) \zeta(\cdot) denotes the Riemann zeta function .

Inspiration .


The answer is 10.

Refaat M. Sayed by Refaat M. Sayed , 24, Egypt

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1 solution

Otto Bretscher
Apr 30, 2016

We have the Taylor Series ln ⁡ ( x + 1 ) x + 1 = ∑ n = 1 ∞ ( − 1 ) n − 1 H n x n \frac{\ln(x+1)}{x+1}=\sum_{n=1}^{\infty}(-1)^{n-1}H_nx^n . Dividing by x x and integrating gives ∑ n = 1 ∞ ( − 1 ) n − 1 H n x n n = ∫ 0 x ln ⁡ ( t + 1 ) t ( t + 1 ) d t = − L i 2 ( − x ) − 1 2 ln ⁡ 2 ( x + 1 ) \sum_{n=1}^{\infty}\frac{(-1)^{n-1}H_n x^n}{n}=\int_{0}^{x}\frac{\ln(t+1)}{t(t+1)}dt=-Li_2(-x)-\frac{1}{2}\ln^2(x+1) . For x = 1 x=1 this comes out to be ζ ( 2 ) 2 − ( ln ⁡ 2 ) 2 2 \frac{\zeta(2)}{2}-\frac{(\ln 2)^2}{2} , so that the answer is 10 \boxed{10}

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