200 Followers Problem (repost)

Calculus Level 5

m = 0 ( 1 ) m 2 m + 1 n = 0 [ ( 1 ) n 4 n + 1 + ( 1 ) n 4 n + 3 ] \large\sum_{m=0}^{\infty} \dfrac{(-1)^m}{2m+1} \sum_{n=0}^{\infty} \left [ \dfrac{(-1)^n}{4n+1}+\dfrac{(-1)^n}{4n+3} \right] If the value of above expression can be expressed in the form π a b c \dfrac{\pi^{a}}{b\sqrt{c}} , find the value of a + b + c a+b+c .


The answer is 12.

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1 solution

Chew-Seong Cheong
Aug 28, 2018

Consider the Maclaurin series of ln ( 1 + z ) \ln (1+z) as follows:

ln ( 1 + z ) = z z 2 2 + z 3 3 z 4 4 + z 5 5 Putting z = i ln ( 1 + i ) = i + 1 2 1 3 i 1 4 + 1 5 i Taking the imaginary part ( ln ( 1 + i ) ) = 1 1 3 + 1 5 1 7 + 1 9 \begin{aligned} \ln (1+z) & = z - \frac {z^2}2 + \frac {z^3}3 - \frac {z^4}4 + \frac {z^5}5 - \cdots & \small \color{#3D99F6} \text{Putting }z = i \\ \ln (1+i) & = i + \frac 12 - \frac 13i - \frac 14 + \frac 15i - \cdots & \small \color{#3D99F6} \text{Taking the imaginary part} \\ \Im (\ln(1+i)) & = 1 - \frac 13 + \frac 15 - \frac 17 + \frac 19 - \cdots \end{aligned}

Therefore,

m = 0 ( 1 ) m 2 m + 1 = ( ln ( 1 + i ) ) = ( ln ( 2 ( cos π 4 + i sin π 4 ) ) ) By Euler’s formula = ( ln ( 2 e π 4 i ) ) = ( 1 2 ln 2 + π 4 i ) = π 4 \begin{aligned} \sum_{m=0}^\infty \frac {(-1)^m}{2m+1} & = \Im (\ln(1+i)) \\ & = \Im \left(\ln \left(\sqrt 2 \left({\color{#3D99F6}\cos \frac \pi 4 +i\sin \frac \pi 4}\right)\right)\right) & \small \color{#3D99F6} \text{By Euler's formula} \\ & = \Im \left(\ln \left(\sqrt 2 {\color{#3D99F6}e^{\frac \pi 4 i}}\right)\right) \\ & = \Im \left(\frac 12 \ln 2 + \frac \pi 4 i\right) \\ & = \frac \pi 4 \end{aligned}

Similarly,

1 2 ln ( 1 + z 1 z ) = z + z 3 3 + z 5 5 + z 7 7 + z 9 9 + Dividing both sides by z 1 2 z ln ( 1 + z 1 z ) = 1 + z 2 3 + z 4 5 + z 6 7 + z 8 9 + Putting z = i 1 2 i ln ( 1 + i 1 i ) = 1 + 1 3 i 1 5 1 7 i + 1 9 + \begin{aligned} \frac 12 \ln \left(\frac {1+z}{1-z}\right) & = z + \frac {z^3}3 + \frac {z^5}5 + \frac {z^7}7 + \frac {z^9}9 + \cdots & \small \color{#3D99F6} \text{Dividing both sides by }z \\ \frac 1{2z} \ln \left(\frac {1+z}{1-z}\right) & = 1 + \frac {z^2}3 + \frac {z^4}5 + \frac {z^6}7 + \frac {z^8}9 + \cdots & \small \color{#3D99F6} \text{Putting }z=\sqrt i \\ \frac 1{2\sqrt i} \ln \left(\frac {1+\sqrt i}{1-\sqrt i}\right) & = 1 + \frac 13i - \frac 15 - \frac 17i + \frac 19 + \cdots \end{aligned}

Implying n = 0 ( ( 1 ) n 4 n + 1 + ( 1 ) n 4 n + 3 ) = ( 1 2 i ln ( 1 + i 1 i ) ) + ( 1 2 i ln ( 1 + i 1 i ) ) \displaystyle \sum_{n=0}^\infty \left(\frac {(-1)^n}{4n+1} + \frac {(-1)^n}{4n+3} \right) = \Re \left(\frac 1{2\sqrt i} \ln \left(\frac {1+\sqrt i}{1-\sqrt i}\right)\right) + \Im \left(\frac 1{2\sqrt i} \ln \left(\frac {1+\sqrt i}{1-\sqrt i}\right)\right)

1 2 i ln ( 1 + i 1 i ) = 1 2 i ln ( ( 1 + i ) ( 1 + i ) ( 1 i ) ( 1 + i ) ) = 1 2 i ln ( ( 1 + 2 i + i ) ( 1 + i ) ( 1 i ) ( 1 + i ) ) = 1 2 i ln ( i + i + i i ) By Euler’s formula = 1 2 i ln ( e π 4 i + e π 2 i + e 3 π 4 i ) = 1 2 i ln ( 1 2 + 0 1 2 + i ( 1 2 + 1 + 1 2 ) ) = 1 2 i ln ( ( 1 + 2 ) i ) = i i ln ( ( 1 + 2 ) e π 2 i ) 2 = ( 1 i ) ( ln ( 1 + 2 ) + π 2 i ) 2 2 = 1 2 2 ( π 2 + ln ( 1 + 2 ) + i ( π 2 ln ( 1 + 2 ) ) ) \begin{aligned} \frac 1{2\sqrt i} \ln \left(\frac {1+\sqrt i}{1-\sqrt i}\right) & = \frac 1{2\sqrt i} \ln \left(\frac {(1+\sqrt i)(1+\sqrt i)}{(1-\sqrt i)(1+\sqrt i)}\right) \\ & = \frac 1{2\sqrt i} \ln \left(\frac {(1+2\sqrt i + i)(1+i)}{(1- i)(1+i)}\right) \\ & = \frac 1{2\sqrt i} \ln \left(\color{#3D99F6}\sqrt i + i + i \sqrt i \right) & \small \color{#3D99F6} \text{By Euler's formula} \\ & = \frac 1{2\sqrt i} \ln \left(\color{#3D99F6}e^{\frac \pi 4 i} + e^{\frac \pi 2 i} + e^{\frac {3\pi}4 i} \right) \\ & = \frac 1{2\sqrt i} \ln \left(\color{#3D99F6}\frac 1{\sqrt 2}+0-\frac 1{\sqrt 2} + i\left(\frac 1{\sqrt 2}+1+\frac 1{\sqrt 2} \right) \right) \\ & = \frac 1{2\sqrt i} \ln \left(\left(1+\sqrt 2 \right)i \right) \\ & = - \frac {i\sqrt i\ln \left(\left(1+\sqrt 2 \right)e^{\frac \pi 2i}\right)}2 \\ & = \frac {(1-i)\left(\ln \left(1+\sqrt 2 \right) + \frac \pi 2i\right)}{2\sqrt 2} \\ & = \frac 1{2\sqrt 2}\left(\frac \pi 2 + \ln(1+\sqrt 2) + i \left(\frac \pi 2 - \ln(1+\sqrt 2) \right)\right) \end{aligned}

Therefore,

n = 0 ( ( 1 ) n 4 n + 1 + ( 1 ) n 4 n + 3 ) = ( 1 2 i ln ( 1 + i 1 i ) ) + ( 1 2 i ln ( 1 + i 1 i ) ) = 1 2 2 ( π 2 + ln ( 1 + 2 ) ) + 1 2 2 ( π 2 ln ( 1 + 2 ) ) = π 2 2 \begin{aligned} \sum_{n=0}^\infty \left(\frac {(-1)^n}{4n+1} + \frac {(-1)^n}{4n+3} \right) & = \Re \left(\frac 1{2\sqrt i} \ln \left(\frac {1+\sqrt i}{1-\sqrt i}\right)\right) + \Im \left(\frac 1{2\sqrt i} \ln \left(\frac {1+\sqrt i}{1-\sqrt i}\right)\right) \\ & = \frac 1{2\sqrt 2}\left(\frac \pi 2 + \ln(1+\sqrt 2) \right) + \frac 1{2\sqrt 2}\left(\frac \pi 2 - \ln(1+\sqrt 2)\right) \\ & = \frac \pi {2\sqrt 2} \end{aligned}

Finally, m = 0 ( 1 ) m 2 m + 1 n = 0 ( ( 1 ) n 4 n + 1 + ( 1 ) n 4 n + 3 ) = π 4 π 2 2 = π 2 8 2 \displaystyle \sum_{m=0}^\infty \frac {(-1)^m}{2m+1} \sum_{n=0}^\infty \left(\frac {(-1)^n}{4n+1} + \frac {(-1)^n}{4n+3} \right) = \frac \pi 4 \cdot \frac \pi {2\sqrt 2} = \frac {\pi^2}{8\sqrt 2} a + b + c = 2 + 8 + 2 = 12 \implies a+b+c = 2+8+2 = \boxed{12} .

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