Can you find the limit of the product of the sum?

Calculus Level 3

σ ( m ) = ∑ k = 0 m a k ρ ( n ) = ∏ m = 1 n ( 1 − σ ( m ) + a σ ( m ) ) \begin{aligned} \displaystyle \sigma(m) &=& \sum_{k=0}^m a^k \\ \displaystyle \rho (n) &=& \prod_{m=1}^n \bigg(1 - \sigma(m) + a \sigma(m) \bigg) \\ \end{aligned}

Let a a be a real number between 0 and 1 exclusive, and denote the functions as described above.

Find lim ⁡ n → ∞ ρ ( n ) \displaystyle \lim_{n\to\infty} \rho(n) .


The answer is 0.

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Jonas Kgomo by Jonas Kgomo , 28, South Africa

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1 solution

Jonas Kgomo
Apr 24, 2014

we compute the sum σ ( m ) = ∑ k = 0 m a k = 1 + a + a 2 + . . . + a m = 1 − a m + 1 1 − a \large \sigma(m)=\sum_{k=0}^ma^k=1+a+a^2+...+a^m=\dfrac{1-a^{m+1}}{1-a}

ρ ( n ) \large \rho(n)

= ∏ m = 1 n ( 1 − σ ( m ) + a σ ( m ) ) = ∏ m = 1 ( 1 − ( 1 − a ) 1 − a m + 1 1 − a ) =\prod_{m=1}^n (1-\sigma(m)+a\sigma(m))=\prod_{m=1}(1-(1-a)\dfrac{1-a^{m+1}}{1-a})

= ∏ m = 1 n a m + 1 = a 2 a 3 a 4 . . . a n + 1 = a n ( n + 1 ) 2 =\prod_{m=1}^na^{m+1}=a^2a^3a^4...a^{n+1}=a^{\dfrac{n(n+1)}{2}}

lim ⁡ n → ∞ ρ ( n ) = 0 = lim ⁡ n → ∞ a n ( n + 1 ) 2 \color{#3D99F6}{\large \lim_{n\to \infty}\rho(n)=0}=\lim_{n\to \infty}a^{\frac{n(n+1)}{2}}

since 0 < a < 1 0<a<1 we see that the limit is 0 0

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