Handle with care II

Calculus Level 3

If the closed form of ∑ m = 1 ∞ ∑ n = 0 ∞ ( − 1 ) n ( 2 m + n ) 2 = π a b − ln ⁡ a a \sum_{m=1}^{\infty} \sum_{n=0}^{\infty} \dfrac{(-1)^n}{(2m+n)^2} = \dfrac{\pi^a}{b} - \dfrac{\ln a}{a } where a a and b b are positive integers. Find the value of a + b a+b .

Source : Math facts, facebook page. The idea to tackle this problem is hinted in Handle with care I .


The answer is 18.

Naren Bhandari by Naren Bhandari , 21, India

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1 solution

S = ∑ m = 1 ∞ ∑ n = 0 ∞ ( − 1 ) n ( 2 m + n ) 2 = 1 2 2 − 1 3 2 + 1 4 2 − 1 5 2 + 1 6 2 − ⋯ for m = 1 + 1 4 2 − 1 5 2 + 1 6 2 − 1 7 2 + 1 8 2 − ⋯ for m = 2 + 1 6 2 − 1 7 2 + 1 8 2 − 1 9 2 + 1 1 0 2 − ⋯ for m = 3 + ⋯ = 1 2 2 + 2 4 2 + 3 6 2 + ⋯ − ( 1 3 2 + 2 5 2 + 3 7 2 + ⋯ ) = ∑ m = 1 ∞ ( 1 4 m − m ( 2 m + 1 ) 2 ) = ∑ m = 1 ∞ ( 1 4 m − 1 2 ( 2 m + 1 ) + 1 2 ( 2 m + 1 ) 2 ) = 1 2 ∑ m = 2 ∞ ( − 1 ) m m + 1 2 ∑ m = 1 ∞ 1 ( 2 m + 1 ) 2 = 1 2 ( 1 − ∑ m = 1 ∞ ( − 1 ) m + 1 m ) + 1 2 ( ∑ m = 0 ∞ 1 ( 2 m + 1 ) 2 − 1 ) = 1 2 ( ∑ m = 0 ∞ 1 ( 2 m + 1 ) 2 − ∑ m = 1 ∞ ( − 1 ) m + 1 m ) = 1 2 ( ∑ m = 1 ∞ 1 m 2 − 1 2 2 ∑ m = 1 ∞ 1 m 2 − ln ⁡ 2 ) = 1 2 ( 3 4 ∑ m = 1 ∞ 1 m 2 − ln ⁡ 2 ) = 1 2 ( 3 4 ζ ( 2 ) − ln ⁡ 2 ) where ζ ( ⋅ ) denotes the Riemann = 1 2 ( 3 4 × π 2 6 − ln ⁡ 2 ) zeta function and ζ ( 2 ) = π 2 6 . = π 2 16 − ln ⁡ 2 2 \begin{aligned} S & = \sum_{m=1}^\infty \sum_{n=0}^\infty \frac {(-1)^n}{(2m+n)^2} \\ & = \frac 1{2^2} - \frac 1{3^2} + \frac 1{4^2} - \frac 1{5^2} + \frac 1{6^2} - \cdots & \small \blue{\text{for }m=1} \\ & \quad + \frac 1{4^2} - \frac 1{5^2} + \frac 1{6^2} - \frac 1{7^2} + \frac 1{8^2} - \cdots & \small \blue{\text{for }m=2} \\ & \quad + \frac 1{6^2} - \frac 1{7^2} + \frac 1{8^2} - \frac 1{9^2} + \frac 1{10^2} - \cdots & \small \blue{\text{for }m=3} \\ & \quad + \cdots \\ & = \frac 1{2^2} + \frac 2{4^2} + \frac 3{6^2} + \cdots - \left(\frac 1{3^2} + \frac 2{5^2} + \frac 3{7^2} + \cdots \right) \\ & = \sum_{m=1}^\infty \left(\frac 1{4m} - \frac m{(2m+1)^2} \right) \\ & = \sum_{m=1}^\infty \left(\frac 1{4m} - \frac 1{2(2m+1)} + \frac 1{2(2m+1)^2} \right) \\ & = \frac 12 \sum_{m=2}^\infty \frac {(-1)^m}{m} + \frac 12 \sum_{m=1}^\infty \frac 1{(2m+1)^2} \\ & = \frac 12 \left(1 - \sum_{m=1}^\infty \frac {(-1)^{m+1}}{m} \right) + \frac 12 \left(\sum_{m=0}^\infty \frac 1{(2m+1)^2} - 1 \right) \\ & = \frac 12 \left(\red{\sum_{m=0}^\infty \frac 1{(2m+1)^2}} - \blue{\sum_{m=1}^\infty \frac {(-1)^{m+1}}{m}} \right) \\ & = \frac 12 \left(\red{\sum_{m=1}^\infty \frac 1{m^2} - \frac 1{2^2}\sum_{m=1}^\infty \frac 1{m^2}} - \blue{\ln 2} \right) \\ & = \frac 12 \left(\red{\frac 34 \sum_{m=1}^\infty \frac 1{m^2}} - \ln 2 \right) \\ & = \frac 12 \left(\red{\frac 34 \zeta (2)} - \ln 2 \right) & \small \red{\text{where }\zeta(\cdot) \text{ denotes the Riemann}} \\ & = \frac 12 \left(\red{\frac 34 \times \frac {\pi^2}6} - \ln 2 \right) & \small \red{\text{zeta function and }\zeta(2) = \frac {\pi^2}6.} \\ & = \frac {\pi^2}{16} - \frac {\ln 2}2 \end{aligned}

Therefore, a + b = 2 + 16 = 18 a+b=2+16 = \boxed{18} .

Reference: Riemann zeta function

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