Handle with care I

Calculus Level 3

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

Notation : G G is Catalan's constant .

This is an original and modified version of Handle with care II . It would be better to solve problem without integration.


The answer is 37.

Naren Bhandari by Naren Bhandari , 21, India

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2 solutions

S = ∑ m = 1 ∞ ∑ n = 0 ∞ ( − 1 ) m ( 2 m + n ) 2 = − 1 2 2 − 1 3 2 − 1 4 2 − 1 5 2 − 1 6 2 − 1 7 2 − ⋯ for m = 1 + 1 4 2 + 1 5 2 + 1 6 2 + 1 7 2 + 1 8 2 + 1 9 2 + ⋯ for m = 2 − 1 6 2 − 1 7 2 − 1 8 2 − 1 9 2 − 1 1 0 2 − 1 1 1 2 − ⋯ for m = 3 + ⋯ = − 1 2 2 − 1 3 2 − 1 6 2 − 1 7 2 − 1 1 0 2 − 1 1 1 2 − ⋯ \begin{aligned} S & = \sum_{m=1}^\infty \sum_{n=0}^\infty \frac {(-1)^m}{(2m+n)^2} \\ & = - \frac 1{2^2} - \frac 1{3^2} - \frac 1{4^2} - \frac 1{5^2} - \frac 1{6^2} - \frac 1{7^2} - \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} + \frac 1{9^2} + \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} - \frac 1{11^2} - \cdots & \small \blue{\text{for } m = 3} \\ & \quad + \cdots \\ & = - \frac 1{2^2} - \frac 1{3^2} - \frac 1{6^2} - \frac 1{7^2} - \frac 1{10^2} - \frac 1{11^2} - \cdots \end{aligned}

= − ( 1 2 2 + 1 6 2 + 1 1 0 2 + ⋯ ) + ( − 1 3 2 − 1 7 2 − 1 1 1 2 − ⋯ ) = − 1 2 2 ( 1 1 2 + 1 3 2 + 1 5 2 + 1 7 2 + ⋯ ) + 1 2 [ ( 1 1 2 − 1 3 2 + 1 5 2 − ⋯ ) − ( 1 1 2 + 1 3 2 + 1 5 2 + ⋯ ) ] = − 1 4 [ ( 1 1 2 + 1 2 2 + 1 3 2 + ⋯ ) − ( 1 2 2 + 1 4 2 + 1 6 2 + ⋯ ) ] + 1 2 [ G − ( 1 1 2 + 1 3 2 + 1 5 2 + ⋯ ) ] where G denotes the Catalan’s constant. = − 1 4 [ ( 1 1 2 + 1 2 2 + 1 3 2 + ⋯ ) − 1 2 2 ( 1 1 2 + 1 2 2 + 1 3 2 + ⋯ ) ] + 1 2 [ G − ( 1 1 2 + 1 3 2 + 1 5 2 + ⋯ ) ] = − 1 4 [ ζ ( 2 ) − 1 4 ζ ( 2 ) ] + 1 2 [ G − 3 4 ζ ( 2 ) ] where ζ ( ⋅ ) denotes the Riemann zeta function = G 2 − 9 ζ ( 2 ) 16 = G 2 − 3 π 2 32 and ζ ( 2 ) = π 2 6 \begin{aligned} \ \ \ & = \small - \blue{\left(\frac 1{2^2} + \frac 1{6^2} + \frac 1{10^2} + \cdots \right)} + \red{\left(-\frac 1{3^2} - \frac 1{7^2} - \frac 1{11^2} - \cdots \right)} \\ & = \small - \blue{\frac 1{2^2} \left(\frac 1{1^2} + \frac 1{3^2} + \frac 1{5^2} + \frac 1{7^2} + \cdots \right)} \\ & \quad \small + \red{\frac 12 \left[\left(\frac 1{1^2} - \frac 1{3^2} + \frac 1{5^2} - \cdots \right)-\left(\frac 1{1^2} + \frac 1{3^2} + \frac 1{5^2} + \cdots \right)\right]} \\ & = \small - \blue{\frac 14 \left[\left(\frac 1{1^2} + \frac 1{2^2} + \frac 1{3^2} + \cdots \right)-\left(\frac 1{2^2} + \frac 1{4^2} + \frac 1{6^2} + \cdots \right)\right]} \\ & \quad \small + \red{\frac 12 \left[G-\left(\frac 1{1^2} + \frac 1{3^2} + \frac 1{5^2} + \cdots \right)\right]} & \small \red{\text{where }G \text{ denotes the Catalan's constant.}} \\ & = \small - \blue{\frac 14 \left[\left(\frac 1{1^2} + \frac 1{2^2} + \frac 1{3^2} + \cdots \right)- \frac 1{2^2} \left(\frac 1{1^2} + \frac 1{2^2} + \frac 1{3^2} + \cdots \right)\right]} \\ & \quad \small + \red{\frac 12 \left[G-\left(\frac 1{1^2} + \frac 1{3^2} + \frac 1{5^2} + \cdots \right)\right]} \\ & = - \blue{\frac 14 \left[\zeta (2) - \frac 14 \zeta(2)\right]} + \red{\frac 12 \left[G-\frac 34 \zeta(2) \right]} & \small \blue {\text{where }\zeta (\cdot) \text{ denotes the Riemann zeta function}} \\ & = \frac G2 - \frac {9\blue{\zeta(2)}}{16} = \frac G2 - \frac {3\pi^2}{32} & \small \blue{\text{and }\zeta(2) = \frac {\pi^2}6} \end{aligned}

Therefore, a + b + c = 2 + 32 + 3 = 37 a+b+c = 2+32+3 = \boxed{37} .

References:

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Naren Bhandari
Oct 6, 2019

Here I wish to share my solution without integration

∑ m = 1 ∞ ∑ n = 0 ∞ ( − 1 ) m ( 2 m + n ) 2 = ∑ m = 1 ∞ ( − 1 ) m ∑ n = 0 ∞ 1 ( 2 m + n ) 2 \sum_{m=1}^{\infty} \sum_{n=0}^{\infty} \dfrac{(-1)^m}{(2m+n)^2}=\sum_{m=1}^{\infty}(-1)^m\sum_{n=0}^{\infty}\dfrac{1}{(2m+n)^2} if we write the inner sum in series form we notice that ∑ n = 0 ∞ 1 ( 2 m + n ) 2 = 1 4 ∑ n = 0 ∞ 1 ( m + n ) 2 ⏟ S 1 + ∑ n = 1 , 3 , 5 ⋯ ∞ 1 ( 2 m + n ) 2 ⏟ S 2 \sum_{n=0}^{\infty} \dfrac{1}{(2m+n)^2}= \underbrace{\dfrac{1}{4}\sum_{n=0}^{\infty} \dfrac{1}{(m+n)^2} }_{\text{}S_1}+\underbrace {\sum_{n=1,3,5\cdots}^{\infty} \dfrac{1}{(2m+n)^2}}_{\text{} S_2} and hence have ∑ m = 1 ∞ ∑ n = 0 ∞ ( − 1 ) m ( 2 m + n ) 2 = ∑ m = 1 ∞ ( − 1 ) m 1 4 ∑ n = 0 ∞ 1 ( m + n ) 2 ⏟ S 1 + ∑ m = 1 ∞ ( − 1 ) m ∑ n = 1 , 3 , 5 ⋯ ∞ 1 ( 2 m + n ) 2 ⏟ S 2 \sum_{m=1}^{\infty} \sum_{n=0}^{\infty} \dfrac{(-1)^m}{(2m+n)^2}=\sum_{m=1}^{\infty}(-1)^m \underbrace{\dfrac{1}{4}\sum_{n=0}^{\infty} \dfrac{1}{(m+n)^2} }_{\text{}S_1}+\sum_{m=1}^{\infty} (-1)^m\underbrace {\sum_{n=1,3,5\cdots }^{\infty} \dfrac{1}{(2m+n)^2}}_{\text{} S_2} ] we now evaluate the S 1 S_1 and S 2 S_2 separately as follows 4 S 1 = ∑ m = 1 ∞ ∑ n = 0 ∞ ( − 1 ) m ( m + n ) 2 = ∑ m = 1 ∞ ( − 1 ) m ( 1 m 2 + 1 ( m + 1 ) 2 ) + ∑ m = 1 ∞ ( − 1 ) m ( 1 ( m + 2 ) 2 + 1 ( m + 3 ) 2 ) + ⋯ = ∑ m = 1 ∞ ( 1 ( m + 1 ) 2 − 1 m 2 ) ⏟ Telescopes + ∑ m = 1 ∞ ( 1 ( m + 3 ) 2 − 1 ( m + 2 ) 2 ) ⏟ Telescopes + ⋯ = − ( 1 1 2 + 1 3 2 + 1 5 2 + ⋯ ) = − ∑ m = 1 ∞ 1 ( 2 m − 1 ) 2 = − ( ∑ m = 1 ∞ 1 m 2 − 1 4 ∑ m = 1 ∞ 1 m 2 ) = − π 2 8 \begin{aligned}4S_1&=\sum_{m=1}^{\infty} \sum_{n=0}^{\infty}\dfrac{(-1)^m}{(m+n)^2}=\sum_{m=1}^{\infty} (-1)^m\left(\dfrac{1}{m^2}+\dfrac{1}{(m+1)^2}\right)\\ &+\sum_{m=1}^{\infty}(-1)^m \left(\dfrac{1}{(m+2)^2}+\dfrac{1}{(m+3)^2}\right)+\cdots \\&=\underbrace{\sum_{m=1}^{\infty} \left(\dfrac{1}{(m+1)^2}-\dfrac{1}{m^2}\right)}_{\text {Telescopes}}+\underbrace{\sum_{m=1}^{\infty} \left(\dfrac{1}{(m+3)^2}-\dfrac{1}{(m+2)^2}\right)}_{\text {Telescopes}}+\cdots \\ &=-\left(\dfrac{1}{1^2}+\dfrac{1}{3^2}+\dfrac{1}{5^2}+\cdots \right) =-\sum_{m=1}^{\infty}\dfrac{1}{(2m-1)^2}\\&=-\left(\sum_{m=1}^{\infty}\dfrac{1}{m^2}-\dfrac{1}{4}\sum_{m=1}^{\infty}\dfrac{1}{m^2}\right)=-\dfrac{\pi^2}{8}\end{aligned} therefore we have S 1 = − π 2 32 S_1=-\frac{\pi^2}{32} and we are left to evaluate S 2 = ∑ m = 1 ∞ ∑ n = 1 , 3 , 5 ⋯ ∞ ( − 1 ) m ( 2 m + n ) 2 = ∑ m = 1 ∞ ( − 1 ) m ( 1 ( 2 m + 1 ) 2 + 1 ( 2 m + 3 ) 2 ) + ∑ m = 1 ∞ ( − 1 ) m ( 1 ( 2 m + 5 ) 2 + 1 ( 2 m + 7 ) 2 ) + ⋯ = ∑ m = 1 ∞ ( 1 ( 2 m + 3 ) 2 − 1 ( 2 m + 1 ) 2 ) ⏟ Telescopes + ∑ m = 1 ∞ ( 1 ( 2 m + 7 ) 2 − 1 ( 2 m + 5 ) 2 ) ⏟ Telescopes + ⋯ = − ( 1 3 2 + 1 7 2 + 1 1 1 2 + ⋯ ) = − ∑ m = 0 ∞ 1 ( 4 m + 3 ) 2 = − 1 4 2 ψ 1 ( 3 4 ) = − 1 16 ( π 2 − 8 G ) \begin{aligned} S_2=& \sum_{m=1}^{\infty}\sum_{n=1,3,5\cdots }^{\infty}\dfrac{(-1)^m}{(2m+n)^2}=\sum_{m=1}^{\infty} (-1)^m\left(\dfrac{1}{(2m+1)^2}+\dfrac{1}{(2m+3)^2}\right)\\ &+\sum_{m=1}^{\infty}(-1)^m \left(\dfrac{1}{(2m+5)^2}+\dfrac{1}{(2m+7)^2}\right)+\cdots \\&=\underbrace{\sum_{m=1}^{\infty} \left(\dfrac{1}{(2m+3)^2}-\dfrac{1}{(2m+1)^2}\right)}_{\text {Telescopes}}+\underbrace{\sum_{m=1}^{\infty} \left(\dfrac{1}{(2m+7)^2}-\dfrac{1}{(2m+5)^2}\right)}_{\text {Telescopes}}+\cdots\\&=-\left(\dfrac{1}{3^2}+\dfrac{1}{7^2}+\dfrac{1}{11^2}+\cdots\right)=-\sum_{m=0}^{\infty}\dfrac{1}{(4m+3)^2}\\&=-\dfrac{1}{4^2} \psi^1\left(\dfrac{3}{4}\right) =-\dfrac{1}{16}\left(\pi^2-8G\right)\end{aligned} or ∑ m = 0 ∞ 1 ( 4 m + 3 ) 2 = ∑ m = 1 ∞ 1 m 2 − ( ∑ m = 1 ∞ 1 ( 2 m ) 2 + ∑ m = 1 ∞ 1 ( 4 m − 3 ) 2 ) = π 2 8 − ( ∑ m = 0 ∞ ( − 1 ) m ( 2 m − 1 ) 2 + ∑ m = 0 ∞ 1 ( 4 m + 3 ) 2 ) ⟹ ∑ m = 0 ∞ 1 ( 4 m + 3 ) 2 = π 2 2 × 8 − G 2 \begin{aligned}\sum_{m=0}^{\infty}\dfrac{1}{(4m+3)^2}&= \sum_{m=1}^{\infty}\dfrac{1}{m^2}-\left(\sum_{m=1}^{\infty}\dfrac{1}{(2m)^2}+\sum_{m=1}^{\infty}\dfrac{1}{(4m-3)^2}\right)\\&= \dfrac{\pi^2}{8}- \left(\sum_{m=0}^{\infty}\dfrac{(-1)^m}{(2m-1)^2}+\sum_{m=0}^{\infty}\dfrac{1}{(4m+3)^2}\right)\\&\implies \sum_{m=0}^{\infty}\dfrac{1}{(4m+3)^2} =\dfrac{\pi^2}{2\times 8}-\dfrac{G}{2} \end{aligned} thus ∑ m = 1 ∞ ∑ n = 0 ∞ ( − 1 ) m ( 2 m + n ) 2 = − π 2 32 + G 2 − π 2 16 = G 2 − 3 π 2 32 \sum_{m=1}^{\infty} \sum_{n=0}^{\infty} \dfrac{(-1)^m}{(2m+n)^2}=-\dfrac{\pi^2}{32}+\dfrac{G}{2}-\dfrac{\pi^2}{16} =\dfrac{G}{2} -\dfrac{3\pi^2}{32}
This solution is of Ali Shather

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