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Calculus Level 5

∑ n = 0 ∞ 240 ( 4 n + 5 ) ( 4 n + 6 ) ( 4 n + 7 ) ( 4 n + 9 ) ( 4 n + 10 ) ( 4 n + 11 ) = a b − c π \sum _{ n=0 }^{ \infty }{ \dfrac { 240 }{ (4n+5)(4n+6)(4n+7)(4n+9)(4n+10)(4n+11) } } =\dfrac { a }{ b } -c\pi

The equation above holds true for positive integers a , b a,b and c c , with a , b a,b coprime. Find a + b + c a+b+c .


The answer is 30.

Hamza A by Hamza A , 19, USA

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3 solutions

S = ∑ n = 0 ∞ 240 ( 4 n + 5 ) ( 4 n + 6 ) ( 4 n + 7 ) ( 4 n + 9 ) ( 4 n + 10 ) ( 4 n + 11 ) = 20 ∑ n = 0 ∞ ( 1 ( 4 n + 5 ) ( 4 n + 9 ) ( 4 n + 10 ) − 1 ( 4 n + 6 ) ( 4 n + 7 ) ( 4 n + 11 ) ) = ∑ n = 0 ∞ ( 1 4 n + 5 − 5 4 n + 9 + 4 4 n + 10 − 4 4 n + 6 + 5 4 n + 7 − 1 4 n + 11 ) = 1 5 − ∑ n = 0 ∞ 4 4 n + 9 − 2 3 + 5 7 + ∑ n = 0 ∞ 4 4 n + 11 = 26 105 − 4 ∑ n = 4 ∞ ( − 1 ) n 2 n + 1 = 26 105 − 4 ( ∑ n = 0 ∞ ( − 1 ) n 2 n + 1 − 1 + 1 3 − 1 5 + 1 7 ) = 22 7 − 4 ∑ n = 0 ∞ ( − 1 ) n 2 n + 1 = 22 7 − 4 tan ⁡ − 1 ( 1 ) = 22 7 − π \begin{aligned} S & = \sum_{n=0}^\infty \frac {240}{(4n+5)(4n+6)(4n+7)(4n+9)(4n+10)(4n+11)} \\ & = 20 \sum_{n=0}^\infty \left( \frac 1{(4n+5)(4n+9)(4n+10)} - \frac 1{(4n+6)(4n+7)(4n+11)} \right) \\ & = \sum_{n=0}^\infty \left( \frac 1{4n+5} - \frac 5{4n+9} + \frac 4{4n+10} - \frac 4{4n+6} + \frac 5{4n+7} - \frac 1{4n+11} \right) \\ & = \frac 15 - \sum_{n=0}^\infty \frac 4{4n+9} - \frac 23 + \frac 57 + \sum_{n=0}^\infty \frac 4{4n+11} \\ & = \frac {26}{105} - 4 \sum_{n=4}^\infty \frac {(-1)^n}{2n+1} \\ & = \frac {26}{105} - 4 \left( \sum_{n=0}^\infty \frac {(-1)^n}{2n+1} - 1 + \frac 13 - \frac 15 + \frac 17 \right) \\ & = \frac {22}7 - 4 \sum _{n=0}^\infty \frac {(-1)^n}{2n+1} \\ & = \frac {22}7 - 4 \tan^{-1} (1) \\ & = \frac {22}7 - \pi \end{aligned}

⟹ a + b + c = 22 + 7 + 1 = 30 \implies a+b+c = 22+7+1 = \boxed{30}

9 Helpful 1 Interesting 2 Brilliant 1 Confused

We just proved that 22/7 is bigger than pi

Stilijan Nanovski - 2 years, 5 months ago
Anirban Karan
Apr 29, 2017

S = ∑ n = 0 ∞ 240 ( 4 n + 5 ) ( 4 n + 6 ) ( 4 n + 7 ) ( 4 n + 9 ) ( 4 n + 10 ) ( 4 n + 11 ) = 240 4 6 ∑ n = 0 ∞ 1 ( n + 5 4 ) ( n + 6 4 ) ( n + 7 4 ) ( n + 9 4 ) ( n + 10 4 ) ( n + 11 4 ) = 240 4 6 ∑ m = 2 ∞ 1 ( m − 3 4 ) ( m − 2 4 ) ( m − 1 4 ) ( m + 1 4 ) ( m + 2 4 ) ( m + 3 4 ) [ taking n = m − 2 ] = 240 4 6 ∑ m = 2 ∞ 1 [ m 2 − ( 3 4 ) 2 ] [ m 2 − ( 1 2 ) 2 ] [ m 2 − ( 1 4 ) 2 ] = 240 4 6 ∑ m = 1 ∞ 1 [ m 2 − ( 3 4 ) 2 ] [ m 2 − ( 1 2 ) 2 ] [ m 2 − ( 1 4 ) 2 ] − 4 21 = − 4 21 + 240 4 6 ∑ m = 1 ∞ [ A m 2 − ( 1 4 ) 2 + B m 2 − ( 1 2 ) 2 + C m 2 − ( 3 4 ) 2 ] [ where A = 32 3 , B = − 256 15 , C = 32 5 ] \begin{aligned} S & = \sum_{n=0}^\infty \frac {240}{(4n+5)(4n+6)(4n+7)(4n+9)(4n+10)(4n+11)} \\ & = \frac {240}{4^6}\sum_{n=0}^\infty \frac{1}{(n+\frac{5}{4})(n+\frac{6}{4})(n+\frac{7}{4})(n+\frac{9}{4})(n+\frac{10}{4})(n+\frac{11}{4})}\\ &=\frac {240}{4^6}\sum_{m=2}^\infty \frac{1}{(m-\frac{3}{4})(m-\frac{2}{4})(m-\frac{1}{4})(m+\frac{1}{4})(m+\frac{2}{4})(m+\frac{3}{4})}\quad \color{#3D99F6}[\text {taking } n=m-2] \\ &=\frac {240}{4^6}\sum_{m=2}^\infty \frac{1}{[m^2-(\frac{3}{4})^2][m^2-(\frac{1}{2})^2][m^2-(\frac{1}{4})^2]}\\ &=\frac {240}{4^6}\sum_{m=1}^\infty \frac{1}{[m^2-(\frac{3}{4})^2][m^2-(\frac{1}{2})^2][m^2-(\frac{1}{4})^2]}-\frac{4}{21}\\ &=-\frac{4}{21}+\frac {240}{4^6}\sum_{m=1}^\infty \Big[\frac{A}{m^2-(\frac{1}{4})^2}+\frac{B}{m^2-(\frac{1}{2})^2}+\frac{C}{m^2-(\frac{3}{4})^2}\Big] \quad \color{#3D99F6}\big[\text{where } A=\frac{32}{3},\, B=-\frac{256}{15},\,C=\frac{32}{5}\big]\\ \end{aligned}

Now, using pole expansion of meromorphic functions, we know, cot ⁡ z = 1 z + 2 z ∑ m = 1 ∞ 1 z 2 − m 2 π 2 \color{#20A900}\cot z=\frac{1}{z}+2z\sum_{m=1}^\infty\frac{1}{z^2-m^2\pi^2} So, taking z = a π z=a\pi , we get ∑ m = 1 ∞ 1 m 2 − a 2 = 1 2 a 2 − π 2 a cot ⁡ ( a π ) \color{#3D99F6}\displaystyle\sum_{m=1}^\infty \frac{1}{m^2-a^2}=\frac{1}{2a^2}-\frac{\pi}{2a}\cot (a\pi)

⟹ S = − 4 21 + 240 4 6 [ A ( 8 − 2 π ) + 2 B + C ( 8 9 + 2 π 3 ) ] = − 4 21 + 10 3 − π [ using values of A , B , C ] = 22 7 − π \begin{aligned}\implies S&=-\frac{4}{21}+\frac {240}{4^6}\Big[A(8-2\pi)+2B+C(\frac{8}{9}+\frac{2\pi}{3})\Big]\\ &=-\frac{4}{21}+\frac{10}{3}-\pi\quad \color{#3D99F6}[\text{using values of }A,B,C]\\ &=\frac{22}{7}-\pi\end{aligned}

So, a + b + c = 22 + 7 + 1 = 30 a+b+c=22+7+1=\boxed{30}

2 Helpful 0 Interesting 0 Brilliant 0 Confused
Syed Shahabudeen
Jun 11, 2020

∑ n = 0 ∞ 240 ( 4 n + 5 ) ( 4 n + 6 ) ( 4 n + 7 ) ( 4 n + 9 ) ( 4 n + 10 ) ( 4 n + 11 ) = 240 Γ ( 7 ) ∑ n = 0 ∞ ( 4 n + 8 ) Γ ( 4 n + 5 ) Γ ( 7 ) Γ ( 4 n + 12 ) = 240 Γ ( 7 ) ∑ n = 0 ∞ ( 4 n + 8 ) ∫ 0 1 t 4 n + 4 ( 1 − t ) 6 d t Γ ( 4 n + 5 ) Γ ( 7 ) Γ ( 4 n + 12 ) = ∫ 0 1 t 4 n + 4 ( 1 − t ) 6 d t = 240 Γ ( 7 ) ∫ 0 1 t 4 ( 1 − t ) 6 ∑ n = 0 ∞ ( 4 n + 8 ) t 4 n d t = 240 Γ ( 7 ) ∫ 0 1 t 4 ( 1 − t ) 6 ( 4 t 4 ( 1 − t 4 ) 2 + 8 ( 1 − t 4 ) ) ∑ n = 0 ∞ ( 4 n + 8 ) t 4 n = 4 t 4 ( 1 − t 4 ) 2 + 8 ( 1 − t 4 ) = 240 Γ ( 7 ) ( 8 ∫ 0 1 ( t 4 ( 1 − t ) 6 ( 1 − t 4 ) ) d t + 4 ∫ 0 1 ( t 8 ( 1 − t ) 6 ( 1 − t 4 ) 2 ) d t ) = 240 Γ ( 7 ) ( 8 ( 18 ln ⁡ 2 − 262 21 ) + 4 ( 1147 42 − 3 π 4 − 36 ln ⁡ 2 ) ) ∫ 0 1 t 4 ( 1 − t ) 6 ( 1 − t 4 ) d t = 18 ln ⁡ 2 − 262 21 ; ∫ 0 1 t 8 ( 1 − t ) 6 ( 1 − t 4 ) 2 = 1147 42 − 3 π 4 − 36 ln ⁡ 2 = 240 6 ! ( 66 7 − 3 π ) = 22 7 − π \begin{aligned} \sum _{ n=0 }^{ \infty }{ \frac { 240 }{ \left( 4n+5 \right) \left( 4n+6 \right) \left( 4n+7 \right) \left( 4n+9 \right) \left( 4n+10 \right) \left( 4n+11 \right) } } &=\frac { 240 }{ \Gamma \left( 7 \right) } \sum _{ n=0 }^{ \infty }{ \frac { \left( 4n+8 \right) \Gamma \left( 4n+5 \right) \Gamma \left( 7 \right) }{ \Gamma \left( 4n+12 \right) } }\\ &= \frac { 240 }{ \Gamma \left( 7 \right) } \sum _{ n=0 }^{ \infty }{ \left( 4n+8 \right) \int _{ 0 }^{ 1 }{ { t }^{ 4n+4 }{ \left( 1-t \right) }^{ 6 }dt } }&&&&&&\textcolor{#BA33D6}{{ \frac { \Gamma \left( 4n+5 \right) \Gamma \left( 7 \right) }{ \Gamma \left( 4n+12 \right) } =\int _{ 0 }^{ 1 }{ { t }^{ 4n+4 }{ \left( 1-t \right) }^{ 6 }dt } }} \\&=\frac { 240 }{ \Gamma \left( 7 \right) } \int _{ 0 }^{ 1 }{ { t }^{ 4 }{ \left( 1-t \right) }^{ 6 }\sum _{ n=0 }^{ \infty }{ { \left( 4n+8 \right) t }^{ 4n } }dt } \\&=\frac { 240 }{ \Gamma \left( 7 \right) } \int _{ 0 }^{ 1 }{ { t }^{ 4 }{ \left( 1-t \right) }^{ 6 }\left( {\frac { 4{ t }^{ 4 } }{ { \left( 1-{ t }^{ 4 } \right) }^{ 2 } } +\frac { 8 }{ \left( 1-{ t }^{ 4 } \right) } } \right) } &&&&&&\textcolor{#EC7300}{{ \sum _{ n=0 }^{ \infty }{ { \left( 4n+8 \right) t }^{ 4n } } ={\frac { 4{ t }^{ 4 } }{ { \left( 1-{ t }^{ 4 } \right) }^{ 2 } } +\frac { 8 }{ \left( 1-{ t }^{ 4 } \right) } }}} \\&=\frac { 240 }{ \Gamma \left( 7 \right) } \left( 8\int _{ 0 }^{ 1 }{ \left( \frac { { t }^{ 4 }{ \left( 1-{ t } \right) }^{ 6 } }{ \left( 1-{ t }^{ 4 } \right) } \right) dt+4\int _{ 0 }^{ 1 }{ \left( \frac { { t }^{ 8 }{ \left( 1-t \right) }^{ 6 } }{ { \left( 1-{ t }^{ 4 } \right) }^{ 2 } } \right) dt } } \right)\\&=\frac { 240 }{ \Gamma \left( 7 \right) } \left( 8\left( 18\ln { 2-\frac { 262 }{ 21 } } \right) +4\left( \frac { 1147 }{ 42 } -\frac { 3\pi }{ 4 } -36\ln { 2 } \right) \right) &&&&&&\textcolor{#20A900}{\int _{ 0 }^{ 1 }{ \frac { { t }^{ 4 }{ \left( 1-{ t } \right) }^{ 6 } }{ \left( 1-{ t }^{ 4 } \right) } dt= } 18\ln { 2-\frac { 262 }{ 21 } } ; \quad\int _{ 0 }^{ 1 }{ \frac { { t }^{ 8 }{ \left( 1-t \right) }^{ 6 } }{ { \left( 1-{ t }^{ 4 } \right) }^{ 2 } } } =\frac { 1147 }{ 42 } -\frac { 3\pi }{ 4 } -36\ln { 2 } }\\&=\frac { 240 }{ 6! } \left( \frac { 66 }{ 7 } -3\pi \right)\\&= \frac { 22 }{ 7 } -\pi \end{aligned} Therefore a + b + c = 22 + 7 + 1 = 30 \boxed{a+b+c=22+7+1=30}

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