A Classical Double Factorial Sum

Calculus Level 5

∑ k = 0 ∞ k ! ( 2 k + 5 ) ! ! \sum _{ k=0 }^{ \infty }{ \frac { k! }{ \left( 2k+5 \right) !! } } If the value of above sum can be expressed in the form A π C B − D E \frac { A{ \pi }^{ C } }{ B } -\frac { D }{ E } such that g c d ( A , B ) = g c d ( D , E ) = 1 gcd\left( A,B \right) =gcd\left( D,E \right) =1 . Then find the value of A + B + C + D + E A+B+C+D+E (This problem was inspired by Haven't seen this before )


The answer is 21.

Syed Shahabudeen by Syed Shahabudeen , 23, India

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1 solution

Syed Shahabudeen
Dec 29, 2019

The following sum can be expressed in the form ∑ k = 3 ∞ k ! ( k − 3 ) ! 2 k ( 2 k ) ! = ∑ k = 3 ∞ ( k ! ) 2 2 k ( 2 k ) ! ( k ) ( k − 1 ) ( k − 2 ) \sum _{ k=3 }^{ \infty }{ \frac { k!\left( k-3 \right) !{ 2 }^{ k } }{ \left( 2k \right) ! } } =\sum _{ k=3 }^{ \infty }{ \frac { { \left( k! \right) }^{ 2 }{ 2 }^{ k } }{ \left( 2k \right) !\left( k \right) \left( k-1 \right) \left( k-2 \right) } } . This can be again written in terms of gamma function ∑ k = 3 ∞ ( k ! ) 2 2 k ( 2 k ) ! ( k ) ( k − 1 ) ( k − 2 ) = ∑ k = 3 ∞ Γ ( k ) Γ ( k + 1 ) 2 k Γ ( 2 k + 1 ) ( k − 1 ) ( k − 2 ) \sum _{ k=3 }^{ \infty }{ \frac { { \left( k! \right) }^{ 2 }{ 2 }^{ k } }{ \left( 2k \right) !\left( k \right) \left( k-1 \right) \left( k-2 \right) } } =\sum _{ k=3 }^{ \infty }{ \frac { \Gamma \left( k \right) \Gamma \left( k+1 \right) { 2 }^{ k } }{ \Gamma \left( 2k+1 \right) \left( k-1 \right) \left( k-2 \right) } } here Γ ( k ) Γ ( k + 1 ) Γ ( 2 k + 1 ) { \frac { \Gamma \left( k \right) \Gamma \left( k+1 \right) }{ \Gamma \left( 2k+1 \right) } } can be expressed in terms of beta integral . Therefore we get ∑ k = 3 ∞ 2 k ( k − 1 ) ( k − 2 ) ∫ 0 1 t k − 1 ( 1 − t ) k d t \sum _{ k=3 }^{ \infty }{ \frac { { 2 }^{ k } }{ \left( k-1 \right) \left( k-2 \right) } \int _{ 0 }^{ 1 }{ { t }^{ k-1 }{ \left( 1-t \right) }^{ k }dt } } on interchanging the sum and integral we get ∫ 0 1 1 t ∑ k = 3 ∞ ( 2 t ( 1 − t ) ) k ( k − 1 ) ( k − 2 ) d t \int _{ 0 }^{ 1 }{ \frac { 1 }{ t } } \sum _{ k=3 }^{ \infty }{ \frac { { \left( 2t\left( 1-t \right) \right) }^{ k } }{ \left( k-1 \right) \left( k-2 \right) } } dt Here the sum of the the series can be obtained from logarthimic series 1 t ∑ k = 3 ∞ ( 2 t ( 1 − t ) ) k ( k − 1 ) ( k − 2 ) = − ( t ) ( 2 ( 1 − t ) ) 2 ln ⁡ ( 1 − 2 t ( 1 − t ) ) + ( t ) ( 2 ( 1 − t ) ) 2 + ( 2 ( 1 − t ) ) ln ⁡ ( 1 − 2 t ( 1 − t ) ) \frac { 1 }{ t } \sum _{ k=3 }^{ \infty }{ \frac { { \left( 2t\left( 1-t \right) \right) }^{ k } }{ \left( k-1 \right) \left( k-2 \right) } } =\quad -\left( t \right) { \left( 2\left( 1-t \right) \right) }^{ 2 }\ln { \left( 1-2t\left( 1-t \right) \right) + } { \left( t \right) \left( 2\left( 1-t \right) \right) }^{ 2 }+\left( 2\left( 1-t \right) \right) \ln { \left( 1-2t\left( 1-t \right) \right) } The remaining part is to just integrate it term by term ( since it would be lengthy I wont be posting it here) ∫ 0 1 ( − ( t ) ( 2 ( 1 − t ) ) 2 ln ⁡ ( 1 − 2 t ( 1 − t ) ) + ( t ) ( 2 ( 1 − t ) ) 2 + ( 2 ( 1 − t ) ) ln ⁡ ( 1 − 2 t ( 1 − t ) ) ) d t = π 6 − 4 9 \int _{ 0 }^{ 1 }{ \left( -\left( t \right) { \left( 2\left( 1-t \right) \right) }^{ 2 }\ln { \left( 1-2t\left( 1-t \right) \right) + } { \left( t \right) \left( 2\left( 1-t \right) \right) }^{ 2 }+\left( 2\left( 1-t \right) \right) \ln { \left( 1-2t\left( 1-t \right) \right) } \right) } dt=\frac { \pi }{ 6 } -\frac { 4 }{ 9 } Therefore as per the question A + B + C + D + E = 21 \boxed{A+B+C+D+E=21}

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