An Unexpected Number 3 (Reuploaded)

Calculus Level 5

There was a mistake in the previous one, so I'm re-uploading it with the mistakes fixed

p n = 6 p n − 1 + ( 2 n − 1 ) 2 p n − 2 p − 1 = 1 , p 0 = 3 q n = 6 q n − 1 + ( 2 n − 1 ) 2 q n − 2 q − 1 = 0 , q 0 = 1 \displaystyle \large \begin{aligned} p_{n} &= 6p_{n-1} + (2n-1)^2p_{n-2} &p_{-1} = 1,\ p_0 = 3 \\ q_{n} &= 6q_{n-1} + (2n-1)^2q_{n-2} &q_{-1} = 0,\ q_0 = 1 \end{aligned}

Find lim ⁡ n → ∞ p n q n \displaystyle \lim_{n \rightarrow \infty} \frac{p_n}{q_n} .

This is a continuation of this problem


The answer is 3.14159265359.

Julian Poon by Julian Poon , 20, Singapore

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2 solutions

Julian Poon
Dec 4, 2018

The answer is indeed π \pi .

Lemma: x p n − 1 + ( 2 n − 1 ) 2 p n − 2 x q n − 1 + ( 2 n − 1 ) 2 q n − 2 = ( 6 + ( 2 n − 1 ) 2 x ) p n − 2 + ( 2 n − 3 ) 2 p n − 3 ( 6 + ( 2 n − 1 ) 2 x ) q n − 2 + ( 2 n − 3 ) 2 q n − 3 \frac{xp_{n-1}+\left(2n-1\right)^2p_{n-2}}{xq_{n-1}+\left(2n-1\right)^2q_{n-2}} = \frac{\left(6+\cfrac{(2n-1)^2}{x}\right)p_{n-2}+(2n-3)^2p_{n-3}}{\left(6+\cfrac{(2n-1)^2}{x}\right)q_{n-2}+\left(2n-3\right)^2q_{n-3}} Proof:

x p n − 1 + ( 2 n − 1 ) 2 p n − 2 x q n − 1 + ( 2 n − 1 ) 2 q n − 2 = x ( 6 p n − 2 + ( 2 n − 3 ) 2 p n − 3 ) + ( 2 n − 1 ) 2 p n − 2 x ( 6 q n − 2 + ( 2 n − 3 ) 2 q n − 3 ) + ( 2 n − 1 ) 2 q n − 2 = ( 6 x + ( 2 n − 1 ) 2 ) p n − 2 + x ( 2 n − 3 ) 2 p n − 3 ( 6 x + ( 2 n − 1 ) 2 ) q n − 2 + x ( 2 n − 3 ) 2 q n − 3 = = ( 6 + ( 2 n − 1 ) 2 x ) p n − 2 + ( 2 n − 3 ) 2 p n − 3 ( 6 + ( 2 n − 1 ) 2 x ) q n − 2 + ( 2 n − 3 ) 2 q n − 3 \begin{aligned} \frac{xp_{n-1}+\left(2n-1\right)^2p_{n-2}}{xq_{n-1}+\left(2n-1\right)^2q_{n-2}} &= \frac{x\left(6p_{n-2}+\left(2n-3\right)^2p_{n-3}\right)+\left(2n-1\right)^2p_{n-2}}{x\left(6q_{n-2}+\left(2n-3\right)^2q_{n-3}\right)+\left(2n-1\right)^2q_{n-2}} \\ &= \frac{\left(6x+\left(2n-1\right)^2\right)p_{n-2}+x\left(2n-3\right)^2p_{n-3}}{\left(6x+\left(2n-1\right)^2\right)q_{n-2}+x\left(2n-3\right)^2q_{n-3}}= \\ &= \frac{\left(6+\cfrac{(2n-1)^2}{x}\right)p_{n-2}+\left(2n-3\right)^2p_{n-3}}{\left(6+\cfrac{(2n-1)^2}{x}\right)q_{n-2}+\left(2n-3\right)^2q_{n-3}} \end{aligned}

p n q n = 6 p n − 1 + ( 2 n − 1 ) 2 p n − 2 6 q n − 1 + ( 2 n − 1 ) 2 q n − 2 = [ 6 + ( 2 n − 1 ) 2 6 ] p n − 2 + ( 2 n − 3 ) 2 p n − 3 [ 6 + ( 2 n − 1 ) 2 6 ] q n − 2 + ( 2 n − 3 ) 2 q n − 3 = [ 6 + ( 2 n − 3 ) 2 6 + ( 2 n − 1 ) 2 6 ] p n − 3 + ( 2 n − 5 ) 2 p n − 4 [ 6 + ( 2 n − 3 ) 2 6 + ( 2 n − 1 ) 2 6 ] q n − 3 + ( 2 n − 5 ) 2 q n − 4 . . . = F n p 0 + 1 2 p − 1 F n q 0 + 1 2 q − 1 = 3 F n + 1 F n = 3 + 1 F n \begin{aligned} \frac{p_n}{q_n} &= \frac{6p_{n-1}+\left(2n-1\right)^2p_{n-2}}{6q_{n-1}+\left(2n-1\right)^2q_{n-2}} \\ &\quad \\ &=\frac{\left[6+\cfrac{(2n-1)^2}{6}\right]p_{n-2}+\left(2n-3\right)^2p_{n-3}}{\left[6+\cfrac{(2n-1)^2}{6}\right]q_{n-2}+\left(2n-3\right)^2q_{n-3}} \\ &\quad \\ &= \frac{\left[6+\cfrac{(2n-3)^2}{6+\cfrac{(2n-1)^2}{6}}\right]p_{n-3}+\left(2n-5\right)^2p_{n-4}}{\left[6+\cfrac{(2n-3)^2}{6+\cfrac{(2n-1)^2}{6}}\right]q_{n-3}+\left(2n-5\right)^2q_{n-4}}\\ &\quad \\ & ... \\ &\quad \\ &=\frac{F_np_0+1^2p_{-1}}{F_nq_0+1^2q_{-1}}=\frac{3F_n+1}{F_n}=3+\frac{1}{F_n} \end{aligned}

Where I've applied the lemma repeatedly and F n = 6 + 3 2 6 + 5 2 6 + 7 2 . . . + . . . 6 + ( 2 n − 1 ) 2 6 \displaystyle F_n=6+\cfrac{3^2}{6+\cfrac{5^2}{6+\cfrac{7^2}{...+\cfrac{...}{6+\cfrac{\left(2n-1\right)^2}{6}}}}}

Hence

lim ⁡ n → ∞ p n q n = lim ⁡ n → ∞ 3 + 1 F n = 3 + 1 2 6 + 3 2 6 + 5 2 6 + 7 2 . . . = π ( ∗ ) \lim_{n \rightarrow \infty} \frac{p_n}{q_n} = \lim_{n \rightarrow \infty} 3+\frac{1}{F_n}=3+\cfrac{1^2}{6+\cfrac{3^2}{6+\cfrac{5^2}{6+\cfrac{7^2}{...}}}} = \pi \quad{(*)}

( ∗ ) (*) is a well known continued fraction for π \pi .

Bonus: By considering the sum ∑ n = 1 ∞ ( − 1 ) n n ( n + 1 ) ( 2 n + 1 ) \displaystyle \sum_{n=1}^{\infty}\frac{(-1)^n}{n(n+1)(2n+1)} , proof ( ∗ ) (*) using Euler's Continued Fraction Formula.

3 Helpful 0 Interesting 1 Brilliant 0 Confused
Otto Bretscher
Dec 5, 2018

We can show, by induction, that q n = ( n + 1 ) ( 2 n + 1 ) ! ! q_n=(n+1)(2n+1)!! , and it is not hard to see that p n − π q n < ( 2 n + 1 ) ! ! p_n-\pi q_n<(2n+1)!! , so that lim ⁡ n → ∞ p n − π q n q n = 0 \lim_{n \to \infty}\frac{p_n-\pi q_n}{q_n}=0 and lim ⁡ n → ∞ p n q n = π \lim_{n \to \infty}\frac{p_n}{q_n}=\pi .

3 Helpful 0 Interesting 1 Brilliant 1 Confused

@Otto Bretscher Wait...what?? Sir, how did you even come to the general formula for the sequence q_n ?????!! I mean, Induction can only be done AFTER we arrive at the formula.........

Aaghaz Mahajan - 2 years, 6 months ago

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If you analyse the ratio r n = q n q n − 1 r_n=\frac{q_n}{q_{n-1}} , as one naturally does in a situation like this, you are led to conjecture that r n = 2 n + 3 + 1 n = ( n + 1 ) ( 2 n + 1 ) n r_n=2n+3+\frac{1}{n}=\frac{(n+1)(2n+1)}{n} ; for example, r 8 = 19.125 , r 9 = 21 + 1 9 , r 10 = 23.1 r_8=19.125, r_9=21+\frac{1}{9}, r_{10}=23.1 . From here you can write and prove the closed form q n = ( n + 1 ) ( 2 n + 1 ) ! ! q_n=(n+1)(2n+1)!! rather easily.

Otto Bretscher - 2 years, 6 months ago

Incredibly elegant, Sir Otto!!

tom engelsman - 6 months, 2 weeks ago

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