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Calculus Level 5

2 ( 1 + ∑ n = 1 ∞ ( 2 n − 1 ) ! ! 2 n ( 2 n + 1 ) n ! ) = 2 ∑ n = 0 ∞ ( − 1 / 2 n ) ( − 1 ) n 2 n + 1 = ? \displaystyle \large 2 \left ( 1 + \sum_{n = 1}^{\infty} \frac{(2n - 1)!!}{2^n (2n + 1)n!} \right ) = 2 \sum_{n = 0}^{\infty} {-1/2 \choose n} \frac{(-1)^n}{2n + 1} =? Note: 2 n n ! = ( 2 n ) ! ! 2^n n! = (2n)!! .


The answer is 3.14159265.

Guillermo Templado by Guillermo Templado , 46, Spain

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2 solutions

Brian Moehring
Aug 15, 2018

Using basic convergence results for power series and the generalized binomial theorem , we have 2 ∑ n = 0 ∞ ( − 1 / 2 n ) ( − 1 ) n 2 n + 1 = 2 ∫ 0 1 ( ∑ n = 0 ∞ ( − 1 / 2 n ) ( − 1 ) n x 2 n ) d x = 2 ∫ 0 1 ( ∑ n = 0 ∞ ( − 1 / 2 n ) ( − x 2 ) n ) d x = 2 ∫ 0 1 ( 1 − x 2 ) − 1 / 2 d x = 2 ∫ 0 π / 2 ( 1 − sin ⁡ 2 θ ) − 1 / 2 ⋅ cos ⁡ θ d θ = 2 ∫ 0 π / 2 d θ = 2 ⋅ π 2 = π ≈ 3.14159 \begin{aligned} 2\sum_{n=0}^\infty \binom{-1/2}{n}\frac{(-1)^n}{2n+1} &= 2\int_0^1 \left(\sum_{n=0}^\infty \binom{-1/2}{n}(-1)^n x^{2n}\right)\,dx \\ &= 2\int_0^1 \left(\sum_{n=0}^\infty \binom{-1/2}{n}(-x^2)^n\right)\,dx \\ &= 2\int_0^1 (1-x^2)^{-1/2}\,dx \\ &= 2\int_0^{\pi/2} (1-\sin^2\theta)^{-1/2} \cdot \cos\theta \, d\theta \\ &= 2\int_0^{\pi/2} d\theta \\ &= 2\cdot \frac{\pi}{2} = \pi \approx \boxed{3.14159} \end{aligned}

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In this proof we are using generalised binomial theorem. We need to justify so many things. If a sequence of functions fn(x) converges uniformly to f(x) in [0,1]. Then integral of limit of fn(x)=limit of integral. Here the limits of integral are from 0 to 1. We have a function (1-x^2)^(-0.5). This can be expanded in power series for -1<x<1. If r<1 then Integral of this function from 0 to r is equal to sum of the integrals of each term of the power series. This we are able to do as the power series converges uniformly in [0,r]. This integral is clearly sininverse(r) we get a series for sininverse of r. We know that that the limit of sininverse(r) exists as r tends to 1- and equals pi/2. We claim that pi/2 is equal to the series we get by substituting r=1. This we can do because of this elementary result. If we have a sequence such that all for each n an is greater than or equal to 0 and a0+a1x+a2x^2+...... converges to the limit A as x tends to 1 from left. Then a0+a1+a2......... converges to A. This is mathematics. We need to justify everything. Life is not easy.

Srikanth Tupurani - 2 years, 6 months ago

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There are many things which need to be justified in case of infinite series. Example we know that l 1/ (1+x) has a power series expansion for x in (-1,1). The power series converges uniformly in the the interval (-r,r)if r in (0,1). We can calculate ln(2)using this. Here we need to use Abel limit theorem.

Srikanth Tupurani - 2 years, 6 months ago
Chew-Seong Cheong
Mar 10, 2018

Consider the sum

S = ∑ n = 1 ∞ ( 2 n − 1 ) ! ! 2 n ( 2 n + 1 ) n ! Note that ( 2 n − 1 ) ! ! = ( 2 n ) ! 2 n n ! = ∑ n = 1 ∞ ( 2 n ) ! 2 2 n ( 2 n + 1 ) ( n ! ) 2 and ( 2 n n ) = ( 2 n ) ! ( n ! ) 2 = ∑ n = 1 ∞ ( 2 n n ) 1 2 2 n ( 2 n + 1 ) \begin{aligned} S & = \sum_{n=1}^\infty \frac {\color{#3D99F6}(2n-1)!!}{2^n(2n+1)n!} & \small \color{#3D99F6} \text{Note that }(2n-1)!! = \frac {(2n)!}{2^n n!} \\ & = \sum_{n=1}^\infty \frac {\color{#3D99F6}(2n)!}{2^{2n}(2n+1)\color{#3D99F6}(n!)^2} & \small \color{#3D99F6} \text{and }\binom {2n}n = \frac {(2n)!}{(n!)^2} \\ & = \sum_{n=1}^\infty {\color{#3D99F6}\binom {2n}n}\frac 1{2^{2n}(2n+1)} \end{aligned}

Using the generating function of central binomial coefficient,

∑ n = 0 ∞ ( 2 n n ) x n = 1 1 − 4 x Replace x with x 2 ∑ n = 0 ∞ ( 2 n n ) x 2 n = 1 1 − 4 x 2 Integrate both sides w.r.t. x . ∑ n = 0 ∞ ( 2 n n ) x 2 n + 1 2 n + 1 = 1 2 sin ⁡ − 1 ( 2 x ) + C where C is the constant of integration. = 1 2 sin ⁡ − 1 ( 2 x ) Since L H S = 0 when x = 0 ⟹ C = 0 ∑ n = 0 ∞ ( 2 n n ) x 2 n 2 n + 1 = sin ⁡ − 1 ( 2 x ) 2 x Divide both sides by x . ∑ n = 0 ∞ ( 2 n n ) 1 2 2 n ( 2 n + 1 ) = sin ⁡ − 1 1 1 = π 2 Put x = 1 2 ∑ n = 1 ∞ ( 2 n n ) 1 2 2 n ( 2 n + 1 ) = π 2 − 1 \begin{aligned} \sum_{n=0}^\infty \binom {2n}n x^n & = \frac 1{\sqrt{1-4x}} & \small \color{#3D99F6} \text{Replace }x \text{ with }x^2 \\ \sum_{n=0}^\infty \binom {2n}n x^{2n} & = \frac 1{\sqrt{1-4x^2}} & \small \color{#3D99F6} \text{Integrate both sides w.r.t. }x. \\ \sum_{n=0}^\infty \binom {2n}n \frac {x^{2n+1}}{2n+1} & = \frac 12 \sin^{-1} (2x) + C & \small \color{#3D99F6} \text{where }C \text{ is the constant of integration.} \\ & = \frac 12 \sin^{-1} (2x) & \small \color{#3D99F6} \text{Since }LHS = 0 \text{ when }x = 0 \implies C = 0 \\ \sum_{n=0}^\infty \binom {2n}n \frac {x^{2n}}{2n+1} & = \frac {\sin^{-1} (2x)}{2x} & \small \color{#3D99F6} \text{Divide both sides by }x. \\ \sum_{\color{#3D99F6}n=0}^\infty \binom {2n}n \frac 1{2^{2n}(2n+1)} & = \frac {\sin^{-1} 1}{1} = \frac \pi 2 & \small \color{#3D99F6} \text{Put }x = \frac 12 \\ \sum_{\color{#D61F06}n=1}^\infty \binom {2n}n \frac 1{2^{2n}(2n+1)} & = \frac \pi 2 - 1 \end{aligned}

Therefore, 2 ( 1 + S ) = 2 ( 1 + π 2 − 1 ) = π ≈ 3.142 2(1+S) = 2\left(1+\dfrac \pi 2 - 1\right) = \pi \approx \boxed{3.142} .

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I understand it

Nahom Assefa - 2 years, 9 months ago

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