But how?

Calculus Level 5

∑ n = 1 ∞ ∑ k = 1 ∞ ∑ i = 1 ∞ n k i ( n + k + i ) ! = A B e \large \displaystyle\sum_{n=1}^\infty\sum_{k=1}^\infty\sum _{ i=1 }^{ \infty }{ \dfrac { nki }{ (n+k+i)! } } =\dfrac { A }{ B } e

The equation above holds true for coprime positive integers A A and B B . Find A + B A+B .

Clarification : e ≈ 2.71828 e \approx 2.71828 is the Euler's number .


The answer is 151.

Hamza A by Hamza A , 19, USA

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2 solutions

Julian Poon
May 25, 2016

I hope there is a better solution to this.

First, notice that the sum above (S) can be expressed as:

S = ∂ ∂ x ∑ a = 1 ∞ ∑ b = 1 ∞ ∑ c = 1 ∞ x a b c ( a + b + c ) ! ∣ x = 1 S=\frac{\partial}{\partial x}\sum _{a=1}^{ \infty}\sum _{b=1}^{ \infty}\sum _{c=1}^{ \infty}\frac{x^{abc}}{\left(a+b+c\right)!}\Bigg\vert_{x=1}

Letting K = a + b + c K=a+b+c

∑ a = 1 ∞ ∑ b = 1 ∞ ∑ c = 1 ∞ x a b c ( a + b + c ) ! = ∑ K = 1 ∞ ∑ a = 1 K − 1 ∑ b = 1 K − 1 − a x a b ( K − a − b ) K ! \sum _{a=1}^{ \infty}\sum _{b=1}^{ \infty}\sum _{c=1}^{ \infty}\frac{x^{abc}}{\left(a+b+c\right)!}=\sum _{K=1}^{\infty }\sum _{a=1}^{K-1}\sum _{b=1}^{K-1-a}\frac{x^{ab(K-a-b)}}{K!}

S = ∂ ∂ x ∑ K = 1 ∞ ∑ a = 1 K − 1 ∑ b = 1 K − 1 − a x a b ( K − a − b ) K ! ∣ x = 1 = ∑ K = 1 ∞ 1 K ! ∑ a = 1 K − 1 ∑ b = 1 K − 1 − a a ⋅ b ⋅ ( K − a − b ) S=\frac{\partial}{\partial x}\sum _{K=1}^{\infty }\sum _{a=1}^{K-1}\sum _{b=1}^{K-1-a}\frac{x^{ab(K-a-b)}}{K!}\Bigg\vert_{x=1}=\sum _{K=1}^{\infty }\frac{1}{K!}\sum _{a=1}^{K-1}\sum _{b=1}^{K-1-a}a\cdot b\cdot \left(K-a-b\right)

Through tedious working,

∑ K = 1 ∞ 1 K ! ∑ a = 1 K − 1 ∑ b = 1 K − 1 − a a ⋅ b ⋅ ( K − a − b ) = 1 120 ∑ K = 1 ∞ 1 K ! ( K − 1 ) ( K ) ( K 3 + K 2 − 4 K − 4 ) = 31 120 e \sum _{K=1}^{\infty }\frac{1}{K!}\sum _{a=1}^{K-1}\sum _{b=1}^{K-1-a}a\cdot b\cdot \left(K-a-b\right)=\frac{1}{120}\sum _{K=1}^{\infty }\frac{1}{K!}\left(K-1\right)\left(K\right)\left(K^3+K^2-4K-4\right)=\frac{31}{120}e

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Moderator note:

It's not clear to me why you needed to use the partial derivative. You could have gone through the same argument directly, and provide a justification for the interchange of order (which is alright since all of the terms are positive).

The "tedious working" part could be justified with some combinatorial counting of ∑ a b c \sum abc given that a + b + c = K a+b+c = K .

Hamza A
May 27, 2016

let S S denote the above sum,

first we do something that'll help us in converting the sum into an integral,we write the following sum as ∑ a , b , c = 1 ∞ a b c ( a + b + c ) ! = ∑ a , b , c = 1 ∞ a b c ( a + b ) ! ( c − 1 ) ! ⋅ ( a + b ) ! ( c − 1 ) ! ( a + b + c ) ! \sum _{ a,b,c=1 }^{ \infty }{ \frac { abc }{ (a+b+c)! } } =\sum _{ a,b,c=1 }^{ \infty }{ \frac { abc }{ (a+b)!(c-1)! } \cdot \frac { (a+b)!(c-1)! }{ (a+b+c)! } }

converting the 2nd fraction into an integral we have

S = ∑ a , b , c = 1 ∞ a b c ( a + b ) ! ( c − 1 ) ! ∫ 0 1 x a + b ( 1 − x ) c − 1 d x S=\sum _{ a,b,c=1 }^{ \infty }{ \frac { abc }{ (a+b)!(c-1)! } \int _{ 0 }^{ 1 }{ { x }^{ a+b }\left( 1-x \right) ^{ c-1 } } dx}

interchanging the summation and integral we get

S = ∫ 0 1 ( ∑ a , b = 1 ∞ a b x a + b ( a + b ) ! ) ( ∑ c = 1 ∞ c ( 1 − x ) c − 1 ( c − 1 ) ! ) d x S=\int _{ 0 }^{ 1 }{ \left( \sum _{ a,b=1 }^{ \infty }{ \frac { ab{ x }^{ a+b } }{ (a+b)! } } \right) \left( \sum _{ c=1 }^{ \infty }{ \frac { c(1-x)^{ c-1 } }{ (c-1)! } } \right) dx }

then after we evaluate each summation,we get

S = ∫ 0 1 e x x 2 ( 1 2 + x 6 ) e 1 − x ( 2 − x ) d x = e ∫ 0 1 ( x 3 6 + x 2 2 ) ( 2 − x ) d x = 31 120 e S= \int _{ 0 }^{ 1 }{ { e }^{ x }{ x }^{ 2 }\left( \frac { 1 }{ 2 } +\frac { x }{ 6 } \right) { e }^{ 1-x }(2-x)dx } =e\int _{ 0 }^{ 1 }{ \left( \frac { { x }^{ 3 } }{ 6 } +\frac { { x }^{ 2 } }{ 2 } \right) (2-x)dx }= \boxed{\frac { 31 }{ 120 } e}

Notation used :

∑ n 1 , n 2 , n 3 . . . , n k = 1 ∞ a n 1 , n 2 , . . n k = ∑ n 1 = 1 ∞ ∑ n 2 = 1 ∞ ⋅ ⋅ ⋅ ∑ n k = 1 ∞ a n 1 , n 2 , . . n k \sum _{ { n }_{ 1 },{ n }_{ 2 },{ n }_{ 3 }...,{ n }_{ k }=1 }^{ \infty }{ { a }_{ { n }_{ 1 },{ n }_{ 2 },..{ n }_{ k } } } =\sum _{ { n }_{ 1 }=1 }^{ \infty }{ \sum _{ { n }_{ 2 }=1 }^{ \infty }{ \cdot \cdot \cdot \sum _{ { n }_{ k }=1 }^{ \infty }{ { a }_{ { n }_{ 1 },{ n }_{ 2 },..{ n }_{ k } } } } }

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