Infinite sum with golden number

Calculus Level 4

Evaluate

S = ∑ m = 1 ∞ ∑ n = 1 ∞ m 2 n φ m ( n φ m + m φ n ) \large S=\sum_{m=1}^{\infty}\sum_{n=1}^{\infty} \frac{m^2n}{\varphi^m\left(n\varphi^m+m\varphi^n\right)}

Notation: φ = 1 + 5 2 \varphi=\dfrac{1+\sqrt5}{2} denotes the golden ratio .


The answer is 8.972.

Khang Nguyen Thanh by Khang Nguyen Thanh , 27, Vietnam

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1 solution

We may swap the order of summation and rename the dummy variables to get S = ∑ m = 1 ∞ ∑ n = 1 ∞ m n 2 φ n ( n φ m + m φ n ) S=\sum_{m=1}^{\infty}\sum_{n=1}^{\infty} \dfrac{mn^2}{\varphi^n\left(n\varphi^m+m\varphi^n\right)} Thus, 2 S = S + S = ∑ m = 1 ∞ ∑ n = 1 ∞ m n n φ m + m φ n ( m φ m + n φ n ) = ∑ m = 1 ∞ ∑ n = 1 ∞ m n φ m φ n = ( ∑ m = 1 ∞ m φ m ) 2 2S=S+S=\sum_{m=1}^{\infty}\sum_{n=1}^{\infty} \dfrac{mn}{n\varphi^m+m\varphi^n}\left(\dfrac{m}{\varphi^m}+\dfrac{n}{\varphi^n}\right)=\sum_{m=1}^{\infty}\sum_{n=1}^{\infty} \dfrac{mn}{\varphi^m\varphi^n}=\left(\sum_{m=1}^{\infty} \dfrac{m}{\varphi^m}\right)^2 For ∣ x ∣ < 1 |x|<1 , we have ∑ n = 1 ∞ n x n = x ⋅ d d x ∑ n = 0 ∞ x n = x ⋅ ( 1 1 − x ) = x ( 1 − x ) 2 \sum_{n=1}^{\infty} nx^n=x\cdot\dfrac{d}{dx} \sum_{n=0}^{\infty} x^n=x\cdot\left(\dfrac{1}{1-x}\right)=\dfrac{x}{(1-x)^2} hence 2 S = ( ∑ n = 1 ∞ n φ − n ) 2 = ( φ − 1 ( 1 − φ − 1 ) 2 ) 2 = φ 2 ( φ − 1 ) 4 2S=\left(\sum_{n=1}^{\infty} n\varphi^{-n}\right)^2=\left(\dfrac{\varphi^{-1}}{\left(1-\varphi^{-1}\right)^2}\right)^2=\dfrac{\varphi^2}{(\varphi-1)^4} so S = φ 2 2 ( φ − 1 ) 4 ≈ 8.972 S=\dfrac{\varphi^2}{2(\varphi-1)^4}\approx \boxed{8.972}

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