Playing with Radicals

Calculus Level 5

P = 1 2 ( 1 2 + 2 1 ) × 1 2 ( 2 3 + 3 2 ) × 1 2 ( 3 4 + 4 3 ) × ⋯ P=\frac{1}{2}\left(\frac{\sqrt{1}}{\sqrt{2}}+\frac{\sqrt{2}}{\sqrt{1}}\right)\times\frac{1}{2}\left(\frac{\sqrt{2}}{\sqrt{3}}+\frac{\sqrt{3}}{\sqrt{2}}\right)\times \frac{1}{2}\left(\frac{\sqrt{3}}{\sqrt{4}}+\frac{\sqrt{4}}{\sqrt{3}}\right)\times \cdots

The infinite product P P can be written as P = a π b P=a\pi^{b} for rational numbers a a and b b . Enter a × b a\times b as your answer.


The answer is -1.

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Otto Bretscher by Otto Bretscher , 65, USA

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3 solutions

Otto Bretscher
Oct 12, 2015

We will use the Wallis Formula , 2 ∗ 2 ∗ 4 ∗ 4 ∗ 6 ∗ 6 ∗ 8 1 ∗ 3 ∗ 3 ∗ 5 ∗ 5 ∗ 7 ∗ 7 . . . . = π 2 \frac{2*2*4*4*6*6*8}{1*3*3*5*5*7*7}....=\frac{\pi}{2}

Now P 2 = ∏ k = 1 ∞ ( 2 k + 1 ) 2 ( 2 k ) ( 2 k + 2 ) = 3 ∗ 3 ∗ 5 ∗ 5 ∗ 7 ∗ 7 2 ∗ 4 ∗ 4 ∗ 6 ∗ 6 ∗ 8 . . . = 4 π P^2=\prod_{k=1}^{\infty}\frac{(2k+1)^2}{(2k)(2k+2)}=\frac{3*3*5*5*7*7}{2*4*4*6*6*8}...=\frac{4}{\pi} so P = 2 π − 1 / 2 P=2\pi^{-1/2} and a b = − 1 ab=\boxed{-1}

15 Helpful 0 Interesting 0 Brilliant 0 Confused

Cool! I didn't know about the Wallis Formula. Yet another thing I've learned from you. :)

Brian Charlesworth - 5 years, 8 months ago

Thanks Calculus Laoshi (literally "old (learned) master" in Chinese, meaning teacher) for the Wallis Formula. It is a real bonus trying your problems.

Chew-Seong Cheong - 5 years, 7 months ago

Looking first at the partial products, we have that

P n = ∏ k = 1 n 1 2 ( k + ( k + 1 ) k ( k + 1 ) ) = ( 2 n + 1 ) ! ! 2 n ∗ n ! ∗ n + 1 = ( 2 n + 1 ) ! 2 2 n ∗ ( n ! ) 2 ∗ n + 1 . P_{n} = \displaystyle\prod_{k=1}^{n} \dfrac{1}{2}\left(\dfrac{k + (k + 1)}{\sqrt{k(k + 1)}}\right) = \dfrac{(2n + 1)!!}{2^{n}*n!*\sqrt{n + 1}} = \dfrac{(2n + 1)!}{2^{2n}*(n!)^{2}*\sqrt{n + 1}}.

Next, we can employ Stirling's Approximation n ! ∼ 2 π n ( n e ) n n! \sim \sqrt{2\pi n}\left(\dfrac{n}{e}\right)^{n} to find that

P = lim ⁡ n → ∞ P n = lim ⁡ n → ∞ 2 π ( 2 n + 1 ) ∗ ( 2 n + 1 ) 2 n + 1 e 2 n + 1 2 π n ∗ n 2 n e 2 n n + 1 = P = \lim_{n \rightarrow \infty} P_{n} = \lim_{n \rightarrow \infty} \dfrac{\sqrt{2\pi (2n + 1)}*\dfrac{(2n + 1)^{2n + 1}}{e^{2n + 1}}}{2\pi n*\dfrac{n^{2n}}{e^{2n}}\sqrt{n + 1}} =

2 e π lim ⁡ n → ∞ n + 1 2 ∗ ( n + 1 2 ) 2 n n + 1 ∗ n 2 n = \dfrac{2}{e\sqrt{\pi}} \lim_{n \rightarrow \infty} \dfrac{\sqrt{n + \frac{1}{2}}*(n + \frac{1}{2})^{2n}}{\sqrt{n + 1}*n^{2n}} =

2 e π lim ⁡ n → ∞ n + 1 2 n + 1 lim ⁡ n → ∞ ( 1 + 1 2 n ) 2 n = 2 e π ∗ 1 ∗ e = 2 π . \dfrac{2}{e\sqrt{\pi}} \lim_{n \rightarrow \infty} \dfrac{\sqrt{n + \frac{1}{2}}}{\sqrt{n + 1}} \lim_{n \rightarrow \infty} \left(1 + \dfrac{1}{2n}\right)^{2n} = \dfrac{2}{e\sqrt{\pi}}*1*e = \dfrac{2}{\sqrt{\pi}}.

Thus a b = 2 ∗ ( − 1 2 ) = − 1 . ab = 2*\left(-\dfrac{1}{2}\right) = \boxed{-1}.

10 Helpful 0 Interesting 0 Brilliant 0 Confused

Very nice (upvote)! It is perhaps a bit "casual" to write that n ! → 2 π n ( n e ) n n! \rightarrow \sqrt{2\pi n}\left(\frac{n}{e}\right)^n .... for the benefit of other members, would you mind clarifying?

Otto Bretscher - 5 years, 8 months ago

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Yes, I suppose it is too "casual", as the approximation is an asymptotic formula. More precisely, Stirling's Formula states that

n ! = 2 π n ( n e ) n ( 1 + O ( 1 n ) ) . n! = \sqrt{2\pi n}\left(\dfrac{n}{e}\right)^{n}\left(1 + O\left(\dfrac{1}{n}\right)\right).

Stirling's formula is in turn the first approximation of the Stirling series, which is technically an asymptotic expansion and not a convergent series. But if we were to let this series be g ( n ) g(n) and the approximation in my solution be f ( n ) f(n) then it is the case that lim ⁡ n → ∞ f ( n ) g ( n ) = 1 , \lim_{n \rightarrow \infty} \dfrac{f(n)}{g(n)} = 1, fitting the definition of an asymptotic formula. So if my solution were to have been less casual, it should have ventured into the realm of asymptotic analysis .

Brian Charlesworth - 5 years, 8 months ago

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Thanks for the detailed explanation! A simple solution is to write n ! ∼ 2 π n ( n e ) n n! \sim \sqrt{2\pi n}\left(\frac{n}{e}\right)^n , meaning that the ratio goes to 1...it's precise without requiring more writing.

Otto Bretscher - 5 years, 8 months ago

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@Otto Bretscher – Great! I've changed the → \rightarrow to ∼ \sim in my solution as you've suggested. Thanks. :)

Brian Charlesworth - 5 years, 8 months ago

Luckily when I hit that "What do U need to solve the problem?" button, I was directed to Sterling's Formula and thanks to it I can solve the problem. Lol.

Gian Sanjaya - 5 years, 8 months ago

Yes, much easier to use Stirling.

Chew-Seong Cheong - 5 years, 8 months ago
Chew-Seong Cheong
Oct 11, 2015

P = lim ⁡ n → ∞ ∏ k = 1 n 1 2 ( k k + 1 + k + 1 k ) = lim ⁡ n → ∞ ∏ k = 1 n 1 2 ( 2 k + 1 k ( k + 1 ) ) = lim ⁡ n → ∞ ∏ k = 1 n k + 1 2 k ( k + 1 ) ∏ k = 1 n ( k + 1 2 ) = 2 Γ ( n + 3 2 ) π = lim ⁡ n → ∞ 2 Γ ( n + 3 2 ) π Γ ( n + 1 ) Γ ( n + 2 ) = 2 π − 1 2 \begin{aligned} P & = \lim_{n \to \infty} \prod_{k=1}^n \frac{1}{2}\left( \frac{\sqrt{k}}{\sqrt{k+1}} + \frac{\sqrt{k+1}}{\sqrt{k}} \right) \\ & = \lim_{n \to \infty} \prod_{k=1}^n \frac{1}{2}\left( \frac{2k+1}{\sqrt{k(k+1)}} \right) \\ & = \lim_{n \to \infty} \prod_{k=1}^n \frac{k+\frac{1}{2}}{\sqrt{k(k+1)}} \quad \quad \quad \quad \quad \quad \quad \quad \small \color{#3D99F6}{\prod_{k=1}^n \left(k+\frac{1}{2}\right) = \frac{2 \Gamma \left(n+\frac{3}{2}\right)}{\sqrt{\pi}}} \\ & = \lim_{n \to \infty} \frac{2 \Gamma \left(n+\frac{3}{2}\right)}{\sqrt{\pi}\sqrt{\Gamma(n+1)\Gamma(n+2)}} \\ & = 2\pi^{-\frac{1}{2}} \end{aligned}

⇒ a × b = 2 ( − 1 2 ) = − 1 \Rightarrow a \times b = 2 \left( -\dfrac{1}{2} \right) = \boxed{-1}

6 Helpful 0 Interesting 0 Brilliant 0 Confused

Very nice (upvote)!

Otto Bretscher - 5 years, 8 months ago

thanks, you always solve in easiest way.

manish kumar singh - 5 years, 7 months ago

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