Do you even telescope? - 2

Calculus Level 3

2 ! + 1 ! 2 ! − 1 ! × 4 ! + 3 ! 4 ! − 3 ! × 6 ! + 5 ! 6 ! − 5 ! × ⋯ \frac{2!+1!}{2!-1!}\times \frac{4!+3!}{4!-3!}\times \frac{6!+5!}{6!-5!}\times \cdots

Find the value of the product above.

Notation :
! ! denotes the factorial notation. For example, 8 ! = 1 × 2 × 3 × ⋯ × 8 8! = 1\times2\times3\times\cdots\times8 .

Hint : Try to express the product using product notation.

Inspiration .

0.5 e e π \pi The product is undefined (or divergent) 1.5 1 1 log ⁡ 3 ( sin ⁡ ( e π ) ) \log_3\left(\sin \left(\frac{e}{\pi}\right)\right) None of these

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Prasun Biswas by Prasun Biswas , 23, India

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3 solutions

Chew-Seong Cheong
Mar 29, 2016

P = 2 ! + 1 ! 2 ! − 1 ! × 4 ! + 3 ! 4 ! − 3 ! × 6 ! + 5 ! 6 ! − 5 ! × . . . = ∏ n = 1 ∞ ( 2 n ) ! + ( 2 n − 1 ) ! ( 2 n ) ! − ( 2 n − 1 ) ! = ∏ n = 1 ∞ ( 2 n − 1 ) ! ( 2 n + 1 ) ( 2 n − 1 ) ! ( 2 n − 1 ) = lim ⁡ m → ∞ ∏ n = 1 m 2 n + 1 2 n − 1 = lim ⁡ m → ∞ 2 m + 1 = ∞ ⇒ The product is undefined. \begin{aligned} P & = \frac{2!+1!}{2!-1!} \times \frac{4!+3!}{4!-3!} \times \frac{6!+5!}{6!-5!} \times ... \\ & = \prod_{n=1}^\infty \frac{(2n)!+(2n-1)!}{(2n)!-(2n-1)!} \\ & = \prod_{n=1}^\infty \frac{(2n-1)!(2n+1)}{(2n-1)!(2n-1)} \\ & = \lim_{m \to \infty} \prod_{n=1}^m \frac{2n+1}{2n-1} \\ & = \lim_{m \to \infty} 2m + 1 \\ & = \infty \quad \Rightarrow \boxed{\text{The product is undefined.}} \end{aligned}

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Can't we should conclude that the product is undefined as each term is >>1 and they are infinite number of terms?

A Former Brilliant Member - 5 years, 2 months ago

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Yes, I was thinking to provide the following solution.

P = lim ⁡ m → ∞ ∏ n = 1 m 2 n + 1 2 n − 1 = lim ⁡ m → ∞ ∏ n = 1 m ( 1 + 2 2 n − 1 ) = ∞ ⇒ The product is undefined. \begin{aligned} P & = \lim_{m \to \infty} \prod_{n=1}^m \frac{2n+1}{2n-1} \\ & = \lim_{m \to \infty} \prod_{n=1}^m \left(1+\frac{2}{2n-1}\right) \\ & = \infty \quad \Rightarrow \boxed{\text{The product is undefined.}} \end{aligned}

Chew-Seong Cheong - 5 years, 2 months ago

Your conclusion is correct but the closed form you get for the partial product is wrong.

∏ n = 1 m 2 n + 1 2 n − 1 = ( 2 m + 1 ) ! ! ( 2 m − 1 ) ! ! = 2 m + 1 ≠ m + 2 \prod_{n=1}^m\frac{2n+1}{2n-1}=\frac{(2m+1)!!}{(2m-1)!!}=2m+1\neq m+2

The product diverges because 2 m + 1 → ∞ 2m+1\to\infty as m → ∞ m\to\infty .

Prasun Biswas - 5 years, 2 months ago

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Thanks, you are right

Chew-Seong Cheong - 5 years, 2 months ago
Fahim Saikat
Apr 1, 2016

2 ! + 1 ! 2 ! − 1 ! × 4 ! + 3 ! 4 ! − 3 ! × 6 ! + 5 ! 6 ! − 5 ! × ⋯ \frac{2!+1!}{2!-1!}\times \frac{4!+3!}{4!-3!}\times \frac{6!+5!}{6!-5!}\times \cdots = 3 × 5 3 × 7 5 × 9 7 × ⋯ 3\times \frac{5}{3}\times \frac{7}{5}\times \frac{9}{7}\times \cdots =2n+1 where n = ∞ n= \infty so, 2 ! + 1 ! 2 ! − 1 ! × 4 ! + 3 ! 4 ! − 3 ! × 6 ! + 5 ! 6 ! − 5 ! × ⋯ = ∞ \frac{2!+1!}{2!-1!}\times \frac{4!+3!}{4!-3!}\times \frac{6!+5!}{6!-5!}\times \cdots =\infty

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. .
Feb 8, 2021

2 ! + 1 ! 2 ! − 1 ! × 4 ! + 3 ! 4 ! − 3 ! × 6 ! + 5 ! 6 ! − 5 ! × ⋯ = 2 + 1 2 − 1 × 24 + 6 24 − 6 × 720 + 120 720 − 120 × ⋯ = 3 1 × 30 18 × 840 600 ⋯ = 3 × 5 3 × 7 5 × ⋯ = 5 × 7 5 × 9 7 × ⋯ = ∞ \frac{2!+1!}{2!-1!} \times \frac{4!+3!}{4!-3!} \times \frac{6!+5!}{6!-5!} \times \cdots = \frac{2+1}{2-1} \times \frac{24+6}{24-6} \times \frac{720+120}{720-120} \times \cdots = \frac{3}{1} \times \frac{30}{18} \times \frac{840}{600} \cdots = 3 \times \frac{5}{3} \times \frac{7}{5} \times \cdots = 5 \times \frac{7}{5} \times \frac{9}{7} \times \cdots = \infty So the answer is ∞ \boxed{ \infty} .

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