A Complicated Binomial Sum

Calculus Level 5

∑ n = 0 ∞ ( 4 n 2 n ) 3 2 − n = A cos ⁡ ( π B ) \large \displaystyle \sum_{n=0}^{\infty} \dbinom{4n}{2n} 32^{-n} = \sqrt{A}\cos\left(\dfrac{\pi}{B}\right)

If the equation above holds true for positive integers A A and B B , then find A × B A\times B .

Notation : ( n k ) = n ! k ! ( n − k ) ! \dbinom nk = \dfrac{n!}{k!(n-k)!} denotes the binomial coefficient .


The answer is 16.

Aditya Narayan Sharma by Aditya Narayan Sharma , 21, India

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2 solutions

Another approach:

Note that for ∣ x ∣ < 1 / 4 |x|<1/4 , the power series expansion of 1 1 − 4 x \frac{1}{\sqrt{1-4x}} is ∑ n ≥ 0 ( 2 n n ) x n \sum_{n\ge 0}\binom{2n}{n}x^n . Thus we get 1 1 − 4 x = ∑ n ≥ 0 ( 2 n n ) x n 1 1 + 4 x = ∑ n ≥ 0 ( 2 n n ) ( − x ) n \frac{1}{\sqrt{1-4x}}=\sum_{n\ge 0}\binom{2n}{n}x^n\\ \frac{1}{\sqrt{1+4x}}=\sum_{n\ge 0}\binom{2n}{n}(-x)^n Adding, we get ∑ n ≥ 0 ( 4 n 2 n ) x 2 n = 1 2 [ 1 1 − 4 x + 1 1 + 4 x ] \sum_{n\ge 0}\binom{4n}{2n}x^{2n}=\frac{1}{2}\left[\frac{1}{\sqrt{1-4x}}+\frac{1}{\sqrt{1+4x}}\right] For our question, x = 1 / 4 2 x=1/4\sqrt{2} . Putting this value in the above equation, and after some calculation, it turns out that the desired sum is 2 cos ⁡ π 8 \boxed{\sqrt{2}\cos\frac{\pi}{8}} .

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Nice way, in general any even power series of f ( x ) f(x) is f ( x ) + f ( − x ) 2 \displaystyle \frac{f(x)+f(-x)}{2} . I didn't calculate putting 1 4 2 \frac{1}{4\sqrt{2}} , so does it yield 2 cos ⁡ π 8 \sqrt{2}\cos\frac{\pi}{8} ? .As you have wriiten 2 cos ⁡ π 8 2\cos\frac{\pi}{8} There will be a 2 \sqrt{2} in place of 2.

Aditya Narayan Sharma - 4 years, 9 months ago

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Oh, thanks for pointing out, correcting the typo.

Samrat Mukhopadhyay - 4 years, 9 months ago

Write ( 4 n 2 n ) = 1 2 π Γ ( n + 1 4 ) Γ ( n + 3 4 ) Γ ( 2 n + 1 ) ( 64 ) n \displaystyle \binom{4n}{2n} = \frac{1}{\sqrt{2}\pi} \frac{\Gamma(n+\frac{1}{4})\Gamma(n+\frac{3}{4})}{\Gamma(2n+1)}(64)^n by Gamma Multiplication Formula.

So we have , S = ∑ n = 0 ∞ ( 4 n 2 n ) 3 2 − n = ∑ n = 0 ∞ 2 n 1 2 π β ( n + 1 4 , n + 3 4 ) \displaystyle S=\sum_{n=0}^{\infty} \binom{4n}{2n}32^{-n} = \sum_{n=0}^{\infty}2^n \frac{1}{\sqrt{2}\pi}\beta(n+\frac{1}{4},n+\frac{3}{4})

Therefore by using integral representation of Beta we have,

S = ∑ n = 0 ∞ 1 2 π ∫ 0 ∞ 2 n x n − 3 4 ( 1 + x ) 2 n + 1 d x = 1 2 π ∫ 0 ∞ x − 3 4 + x 1 4 x 2 + 1 d x \displaystyle S=\sum_{n=0}^{\infty} \frac{1}{\sqrt{2}\pi} \int_{0}^{\infty}2^n \frac{x^{n-\frac{3}{4}}}{(1+x)^{2n+1}} dx =\frac{1}{\sqrt{2}\pi} \int_{0}^{\infty} \frac{x^{-\frac{3}{4}}+x^{\frac{1}{4}}}{x^2+1} dx

Now, Consider generating function of Chebyshev polynomials of second kind

∑ n = 0 ∞ Γ ( n + 1 ) U n ( x ) ( − y ) n Γ ( n + 1 ) = y y 2 + 2 x y + 1 \displaystyle \sum_{n=0}^{\infty} \color{#D61F06}{\Gamma(n+1)}{\rm U}_n(x)\frac{(-y)^n}{\color{#D61F06}{\Gamma(n+1)}} = \frac{y}{y^2+2xy+1}

Using Ramanujan's Master theorem ,

F ( a , s ) = ∫ 0 ∞ x s x 2 + 2 a x + 1 d x = Γ ( s ) Γ ( 1 − s ) U − s ( a ) = π sin ⁡ ( ( 1 − s ) cos ⁡ − 1 a ) sin ⁡ ( π ∣ s ∣ ) 1 − a 2 \displaystyle F(a,s)=\int_{0}^{\infty} \frac{x^s}{x^2+2ax+1}dx = \Gamma(s)\Gamma(1-s)U_{-s}(a) = \frac{\pi\sin((1-s)\cos^{-1}a)}{\sin(\pi |s|)\sqrt{1-a^2}}

Now our sum is equal to : S = 1 2 π ( F ( 0 , − 3 4 ) + F ( 0 , 1 4 ) ) \displaystyle S=\frac{1}{\sqrt{2}\pi}(F(0,\frac{-3}{4})+F(0,\frac{1}{4})) .

After some calculations it comes out to be S = sin ⁡ π 8 + cos ⁡ π 8 = 2 cos ⁡ π 8 \displaystyle S=\sin\frac{\pi}{8}+ \cos\frac{\pi}{8} = \sqrt{2}\cos\frac{\pi}{8}

Thus the answer is 2 × 8 = 16 \boxed{2\times8=16}

8 Helpful 0 Interesting 0 Brilliant 0 Confused

Or you could use contour integration around a semicircular contour in the upper half plane, centred at the origin (with a small semicircle removed to avoid the origin itself), to show that F ( 0 , s ) = ∫ 0 ∞ x s x 2 + 1 d x = 1 2 π sec ⁡ ( 1 2 π s ) − 1 < s < 1 . F(0,s) \; = \; \int_0^\infty \frac{x^s}{x^2+1}\,dx \; = \; \tfrac12\pi \sec\big(\tfrac12\pi s\big) \qquad \qquad -1 < s < 1\;.

Mark Hennings - 4 years, 9 months ago

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Yes its always great to see contour methods from you :). Well I thought of directly implementing the identity,

( 4 n 2 n ) = ∫ ∣ z ∣ = 1 ( 1 + z ) 4 n z 2 n + 1 d z \displaystyle \binom{4n}{2n}=\int_{|z|=1}^{} \frac{(1+z)^{4n}}{z^{2n+1}} dz and then on summing it equals,

∑ n = 0 ∞ ( 4 n 2 n ) x n = ∫ ∣ z ∣ = 1 z z 2 − x ( 1 + z ) 4 d z \displaystyle \sum_{n=0}^{\infty} \binom{4n}{2n}x^n = \int_{|z|=1}^{} \frac{z}{z^2-x(1+z)^4} dz

f ( z ) f(z) tends to infinity on z z attaing values ± 1 ± 1 − 4 x 2 x − 1 \frac{\pm 1 \pm \sqrt{1-4\sqrt{x}}}{2\sqrt{x}}-1 .

Hence on identifying the values which lies inside the unit circle and summing the residues we would get ultimately,

∑ n = 0 ∞ ( 4 n 2 n ) x n = 1 − 16 x + 1 2 − 32 x \displaystyle \sum_{n=0}^{\infty} \binom{4n}{2n}x^n = \frac{\sqrt{\sqrt{1-16x}+1}}{\sqrt{2-32x}} for ∣ x ∣ < 1 16 |x|<\frac{1}{16} .

Aditya Narayan Sharma - 4 years, 9 months ago

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Indeed. Yet another nice approach! I fixed the LaTeX typo in your last formula.

Mark Hennings - 4 years, 9 months ago

This reminds me of the generating function of ( 3 n n ) \dbinom{3n}{n} from Summation Contest.

Ishan Singh - 4 years, 9 months ago

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