400 Followers Problem - Reciprocal Summations 1

Calculus Level 5

S = 1 + 1 3 1 5 1 7 + 1 9 + 1 11 1 13 1 15 + \large{S = 1 + \dfrac{1}{3} - \dfrac{1}{5} - \dfrac{1}{7} + \dfrac{1}{9} + \dfrac{1}{11} - \dfrac{1}{13} - \dfrac{1}{15} + \ldots}

If S S can be expressed as π A B \dfrac{\pi^A}{\sqrt{B}} for positive integers A , B A,B , submit the value of A + B A+B as your answer.


Also try these:
400 Followers Problem - Reciprocal Summations #2
400 Followers Problem - Reciprocal Summations #3


The answer is 9.

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2 solutions

Chew-Seong Cheong
Sep 23, 2015

We note that:

ln ( 1 + x 1 x ) = 2 x + 2 x 3 3 + 2 x 5 5 + 2 x 7 7 + 2 x 9 9 + 2 x 11 11 + 2 x 13 13 + . . . for x < 1 ln ( 1 + e i π 4 1 e i π 4 ) = 2 e i π 4 + 2 e i 3 π 4 3 + 2 e i 5 π 4 5 + 2 e i 7 π 4 7 + 2 e i 9 π 4 9 + 2 e i 11 π 4 11 + 2 e i 13 π 4 13 + . . . [ ln ( 1 + e i π 4 1 e i π 4 ) ] = 2 sin π 4 + 2 sin 3 π 4 3 + 2 sin 5 π 4 5 + 2 sin 7 π 4 7 + 2 sin 9 π 4 9 + 2 sin 11 π 4 11 + . . . = 2 ( 1 + 1 3 1 5 1 7 + 1 9 + 1 11 1 13 1 15 + . . . ) = 2 S S = 1 2 [ ln ( 1 + e i π 4 1 e i π 4 ) ] = 1 2 [ ln ( 1 + 1 2 + i 1 2 1 1 2 i 1 2 ) ] = 1 2 [ ln ( 2 + 1 + i 2 1 i ) ] = 1 2 [ ln ( i ( 2 + 1 ) ) ] = 1 2 [ ln ( 2 + 1 ) + ln ( i ) ] = 1 2 [ ln ( 2 + 1 ) + i π 2 ] = 1 2 × π 2 = π 8 A + B = 9 \begin{aligned} \ln \left(\dfrac{1+x}{1-x} \right) & = 2x + \dfrac{2x^3}{3} + \dfrac{2x^5}{5} + \dfrac{2x^7}{7} + \dfrac{2x^9}{9} + \dfrac{2x^{11}}{11} + \dfrac{2x^{13}}{13} + ... \quad \text{for } |x| <1 \\ \ln \left(\dfrac{1+e^{i\frac{\pi}{4}}}{1-e^{i\frac{\pi}{4}}} \right) & = 2e^{i\frac{\pi}{4}} + \dfrac{2e^{i\frac{3\pi}{4}}}{3} + \dfrac{2e^{i\frac{5\pi}{4}}}{5} + \dfrac{2e^{i\frac{7\pi}{4}}}{7} + \dfrac{2e^{i\frac{9\pi}{4}}}{9} + \dfrac{2e^{i\frac{11\pi}{4}}}{11} + \dfrac{2e^{i\frac{13\pi}{4}}}{13} + ... \\ \Im \left[ \ln \left(\dfrac{1+e^{i\frac{\pi}{4}}}{1-e^{i\frac{\pi}{4}}} \right) \right] & = 2\sin \frac{\pi}{4} + \dfrac{2\sin \frac{3\pi}{4}}{3} + \dfrac{2\sin \frac{5\pi}{4}}{5} + \dfrac{2\sin \frac{7\pi}{4}}{7} + \dfrac{2\sin \frac{9\pi}{4}}{9} + \dfrac{2\sin \frac{11\pi}{4}}{11} + ... \\ & = \sqrt{2} \left(1 + \dfrac{1}{3} - \dfrac{1}{5} - \dfrac{1}{7} + \dfrac{1}{9} + \dfrac{1}{11} - \dfrac{1}{13} - \dfrac{1}{15} +... \right) \\ & = \sqrt{2} S \\ \Rightarrow S & = \frac{1}{\sqrt{2}} \Im \left[ \ln \left(\dfrac{1+e^{i\frac{\pi}{4}}}{1-e^{i\frac{\pi}{4}}} \right) \right] \\ & = \frac{1}{\sqrt{2}} \Im \left[ \ln \left(\frac{1+\frac{1}{\sqrt{2}}+i\frac{1}{\sqrt{2}}}{1-\frac{1}{\sqrt{2}}-i\frac{1}{\sqrt{2}}} \right) \right] \\ & = \frac{1}{\sqrt{2}} \Im \left[ \ln \left(\frac{\sqrt{2}+1+i}{\sqrt{2}-1-i} \right) \right] \\ & = \frac{1}{\sqrt{2}} \Im \left[ \ln \left(i(\sqrt{2}+1) \right) \right] \\ & = \frac{1}{\sqrt{2}} \Im \left[ \ln (\sqrt{2}+1) + \ln (i) \right] \\ & = \frac{1}{\sqrt{2}} \Im \left[ \ln (\sqrt{2}+1) + i\frac{\pi}{2} \right] \\ & = \frac{1}{\sqrt{2}} \times \frac{\pi}{2} = \frac{\pi}{\sqrt{8}} \\ \\ \Rightarrow A + B & = \boxed{9} \end{aligned}

It should be |x| < 1 and not -1 < x < 1. because x= e^(i pi/4) is not a real number, so it doesn't fit the criteria.

Pi Han Goh - 5 years, 8 months ago

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Yes. my mistake. Thanks.

Chew-Seong Cheong - 5 years, 8 months ago
Ronak Agarwal
Dec 29, 2015

The summation is equal to :

S = r = 1 ( 1 ) r 4 r + 1 + ( 1 ) r 4 r + 3 = 0 1 1 1 + x 4 + x 2 1 + x 4 d x \displaystyle S = \sum _{ r=1 }^{ \infty }{ \frac { { (-1) }^{ r } }{ 4r+1 } +\frac { { (-1) }^{ r } }{ 4r+3 } } =\int _{ 0 }^{ 1 }{ \frac { 1 }{ 1+{ x }^{ 4 } } +\frac { { x }^{ 2 } }{ 1+{ x }^{ 4 } } dx }

Put x 1 x = y x-\dfrac{1}{x}=y to get :

S = 0 d y y 2 + 2 = 1 2 π 2 = π 8 \displaystyle S = \int _{ -\infty }^{ 0 }{ \frac { dy }{ { y }^{ 2 }+2 } } =\frac { 1 }{ \sqrt { 2 } } \frac { \pi }{ 2 } =\frac { \pi }{ \sqrt { 8 } }

How did you go from the sum to the integral?

Zach Abueg - 3 years, 10 months ago

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