Even harmonic?

Calculus Level 5

∑ n = 1 ∞ ( − 1 ) n − 1 H 2 n n = A π B C − ( ln ⁡ B ) 2 D \large \sum_{n=1}^\infty (-1)^{n-1} \dfrac{H_{2n}}n = \dfrac{A\pi^B}C - \dfrac{(\ln B)^2}D

If the equation above holds true for positive integers A , B , C A,B,C and D D with A , C A,C coprime, find A + B + C + D A+B+C+D .

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 59.

Aditya Narayan Sharma by Aditya Narayan Sharma , 21, India

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1 solution

Since, ∑ n ≥ 1 H n n x n = 1 2 ln ⁡ 2 ( 1 − x ) + L i 2 ( x ) \displaystyle \sum_{n\ge 1} \frac{H_n}{n}x^n=\frac{1}{2}\ln^2 (1-x)+Li_2(x)

Set x = i x=i & we have Re ( ∑ n ≥ 1 H n n i n ) = ∑ n ≥ 1 ( − 1 ) n H 2 n 2 n \displaystyle \text{Re}(\sum_{n\ge 1} \frac{H_n}{n}i^n) = \sum_{n\ge 1} (-1)^{n}\frac{H_{2n}}{2n}

We are left with ∑ n ≥ 1 ( − 1 ) n H 2 n 2 n = Re ( 1 2 ln ⁡ 2 ( 1 − i ) + L i 2 ( i ) ) \displaystyle \sum_{n\ge 1} (-1)^{n}\frac{H_{2n}}{2n} = \text{Re}(\frac{1}{2}\ln^2 (1-i) + Li_2(i) )

Putting the values successively we have , ∑ n ≥ 1 ( − 1 ) n − 1 H 2 n n = 5 π 2 48 − ln ⁡ 2 2 4 \displaystyle \sum_{n\ge 1} (-1)^{n-1}\frac{H_{2n}}{n} = \frac{5\pi^2}{48}-\frac{\ln^2 2}{4}

3 Helpful 0 Interesting 0 Brilliant 0 Confused

Since H n n \frac{H_n}{n} is a decreasing sequence converging to 0 0 , and since ∣ ∑ n = 1 N e i n θ ∣ ≤ c o s e c 1 2 θ \left| \sum_{n=1}^N e^{in\theta}\right| \; \le \; \mathrm{cosec}\tfrac12\theta for all N ≥ 1 N \ge 1 and all 0 < θ < 2 π 0 < \theta < 2\pi , we (using Dirichlet's Test) deduce that ∑ n = 1 ∞ H n n z n \sum_{n=1}^\infty \frac{H_n}{n}z^n converges for all ∣ z ∣ = 1 |z| =1 except z = 1 z=1 . Thus, using Abel's Theorem, ∑ n = 1 ∞ H n n i n = lim ⁡ r → 1 − ∑ n = 1 ∞ H n n ( i r ) n = lim ⁡ r → 1 − [ 1 2 ln ⁡ ( 1 − i r ) 2 + L i 2 ( i r ) ] = 1 2 ln ⁡ ( 1 − i ) 2 + L i 2 ( i ) \sum_{n=1}^\infty \frac{H_n}{n}i^n \; = \; \lim_{r \to 1-} \sum_{n=1}^\infty \frac{H_n}{n}(ir)^n \; = \; \lim_{r \to 1-}\big[\tfrac12\ln(1-ir)^2 + \mathrm{Li}_2(ir)\big] \; = \; \tfrac12\ln(1-i)^2 + \mathrm{Li}_2(i) and we are done.

Mark Hennings - 4 years, 10 months ago

Taking time out to show that the series ∑ n H n i n n \sum_n \frac{H_n i^n}{n} converges, and using Abel's Theorem, so that we can take the limit of the first equation as x → i x \to i radially outwards.

Mark Hennings - 4 years, 10 months ago

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Yes btw can you please elaborate the convergence here ? I'm weak at that.

Aditya Narayan Sharma - 4 years, 10 months ago

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