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Question: $S = \sum_{n=0}^{\infty} \frac{1 \cdot 5 \cdot \dots \cdot (4n+1)}{3 \cdot 7 \cdot \dots \cdot (4n+...

S=∑n=0∞1⋅5⋅⋯⋅(4n+1)3⋅7⋅⋯⋅(4n+3)S = \sum_{n=0}^{\infty} \frac{1 \cdot 5 \cdot \dots \cdot (4n+1)}{3 \cdot 7 \cdot \dots \cdot (4n+3)}

A

Γ(3/4)Γ(1/2)Γ(1/4)\frac{\Gamma(3/4)}{\Gamma(1/2)\Gamma(1/4)}

B

Γ(1/4)Γ(3/4)Γ(1/2)\frac{\Gamma(1/4)}{\Gamma(3/4)\Gamma(1/2)}

C

Γ(1/2)Γ(1/4)Γ(3/4)\frac{\Gamma(1/2)}{\Gamma(1/4)\Gamma(3/4)}

D

Γ(1/4)Γ(3/4)Γ(1/2)\frac{\Gamma(1/4)\Gamma(3/4)}{\Gamma(1/2)}

Answer

Γ(3/4)Γ(1/2)Γ(1/4)\frac{\Gamma(3/4)}{\Gamma(1/2)\Gamma(1/4)}

Explanation

Solution

The given series is S=∑n=0∞1⋅5⋅⋯⋅(4n+1)3⋅7⋅⋯⋅(4n+3)S = \sum_{n=0}^{\infty} \frac{1 \cdot 5 \cdot \dots \cdot (4n+1)}{3 \cdot 7 \cdot \dots \cdot (4n+3)}.

We can express the terms of the series using Pochhammer symbols: The numerator is 1⋅5⋅⋯⋅(4n+1)=4n+1(14)n+11 \cdot 5 \cdot \dots \cdot (4n+1) = 4^{n+1} (\frac{1}{4})_{n+1}. The denominator is 3⋅7⋅⋯⋅(4n+3)=4n+1(34)n+13 \cdot 7 \cdot \dots \cdot (4n+3) = 4^{n+1} (\frac{3}{4})_{n+1}.

So, the general term TnT_n is: Tn=4n+1(14)n+14n+1(34)n+1=(14)n+1(34)n+1T_n = \frac{4^{n+1} (\frac{1}{4})_{n+1}}{4^{n+1} (\frac{3}{4})_{n+1}} = \frac{(\frac{1}{4})_{n+1}}{(\frac{3}{4})_{n+1}}.

The series can be written as S=∑n=0∞(14)n+1(34)n+1S = \sum_{n=0}^{\infty} \frac{(\frac{1}{4})_{n+1}}{(\frac{3}{4})_{n+1}}. Let m=n+1m = n+1. Then S=∑m=1∞(14)m(34)mS = \sum_{m=1}^{\infty} \frac{(\frac{1}{4})_{m}}{(\frac{3}{4})_{m}}.

This series can be related to the Gauss hypergeometric function 2F1(a,b;c;z)=∑n=0∞(a)n(b)n(c)nn!zn_2F_1(a,b;c;z) = \sum_{n=0}^{\infty} \frac{(a)_n (b)_n}{(c)_n n!} z^n. A related form is 2F1(a,1;c;z)=∑n=0∞(a)n(c)nzn_2F_1(a,1;c;z) = \sum_{n=0}^{\infty} \frac{(a)_n}{(c)_n} z^n.

Let's consider the series 2F1(a,b;c;1)_2F_1(a,b;c;1). The sum is given by Gauss's summation theorem: 2F1(a,b;c;1)=Γ(c)Γ(c−a−b)Γ(c−a)Γ(c−b)_2F_1(a,b;c;1) = \frac{\Gamma(c)\Gamma(c-a-b)}{\Gamma(c-a)\Gamma(c-b)}, provided Re(c−a−b)>0Re(c-a-b) > 0.

Our series is S=∑n=0∞(14)n+1(34)n+1S = \sum_{n=0}^{\infty} \frac{(\frac{1}{4})_{n+1}}{(\frac{3}{4})_{n+1}}. This can be rewritten as: S=∑n=0∞14(14+1)…(14+n)34(34+1)…(34+n)=∑n=0∞14(54)n34(74)nS = \sum_{n=0}^{\infty} \frac{\frac{1}{4}(\frac{1}{4}+1)\dots(\frac{1}{4}+n)}{\frac{3}{4}(\frac{3}{4}+1)\dots(\frac{3}{4}+n)} = \sum_{n=0}^{\infty} \frac{\frac{1}{4}(\frac{5}{4})_n}{\frac{3}{4}(\frac{7}{4})_n}. This is not directly a 2F1_2F_1 series.

Let's consider the identity: 2F1(a,b;c;z)=Γ(c)Γ(b)Γ(c−b)∫01tb−1(1−t)c−b−1(1−zt)a−1dt_2F_1(a,b;c;z) = \frac{\Gamma(c)}{\Gamma(b)\Gamma(c-b)} \int_0^1 t^{b-1} (1-t)^{c-b-1} (1-zt)^{a-1} dt. Let a=14a = \frac{1}{4}, b=12b = \frac{1}{2}, c=34c = \frac{3}{4}, z=1z = 1. Then 2F1(14,12;34;1)=Γ(34)Γ(12)Γ(34−12)∫01t12−1(1−t)34−12(1−1⋅t)14−1dt_2F_1(\frac{1}{4}, \frac{1}{2}; \frac{3}{4}; 1) = \frac{\Gamma(\frac{3}{4})}{\Gamma(\frac{1}{2})\Gamma(\frac{3}{4}-\frac{1}{2})} \int_0^1 t^{\frac{1}{2}-1} (1-t)^{\frac{3}{4}-\frac{1}{2}} (1-1 \cdot t)^{\frac{1}{4}-1} dt. =Γ(34)Γ(12)Γ(14)∫01t−12(1−t)14(1−t)−34dt= \frac{\Gamma(\frac{3}{4})}{\Gamma(\frac{1}{2})\Gamma(\frac{1}{4})} \int_0^1 t^{-\frac{1}{2}} (1-t)^{\frac{1}{4}} (1-t)^{-\frac{3}{4}} dt. =Γ(34)Γ(12)Γ(14)∫01t−12(1−t)−12dt= \frac{\Gamma(\frac{3}{4})}{\Gamma(\frac{1}{2})\Gamma(\frac{1}{4})} \int_0^1 t^{-\frac{1}{2}} (1-t)^{-\frac{1}{2}} dt. The integral is the Beta function B(12,12)=Γ(12)Γ(12)Γ(12+12)=Γ(12)2Γ(1)=Γ(12)2B(\frac{1}{2}, \frac{1}{2}) = \frac{\Gamma(\frac{1}{2})\Gamma(\frac{1}{2})}{\Gamma(\frac{1}{2}+\frac{1}{2})} = \frac{\Gamma(\frac{1}{2})^2}{\Gamma(1)} = \Gamma(\frac{1}{2})^2. So, 2F1(14,12;34;1)=Γ(34)Γ(12)Γ(14)Γ(12)2=Γ(34)Γ(12)Γ(14)_2F_1(\frac{1}{4}, \frac{1}{2}; \frac{3}{4}; 1) = \frac{\Gamma(\frac{3}{4})}{\Gamma(\frac{1}{2})\Gamma(\frac{1}{4})} \Gamma(\frac{1}{2})^2 = \frac{\Gamma(\frac{3}{4})\Gamma(\frac{1}{2})}{\Gamma(\frac{1}{4})}.

The original series is S=∑n=0∞(1/4)n+1(3/4)n+1S = \sum_{n=0}^{\infty} \frac{(1/4)_{n+1}}{(3/4)_{n+1}}. This series is equal to 2F1(14,1;34;1)−1_2F_1(\frac{1}{4}, 1; \frac{3}{4}; 1) - 1. However, this is not correct.

The series S=∑n=0∞(a)n(b)nS = \sum_{n=0}^{\infty} \frac{(a)_n}{(b)_n} is related to 2F1(a,1;b;1)_2F_1(a,1;b;1). Our series is S=∑n=0∞(1/4)n+1(3/4)n+1S = \sum_{n=0}^{\infty} \frac{(1/4)_{n+1}}{(3/4)_{n+1}}. Let m=n+1m=n+1. S=∑m=1∞(1/4)m(3/4)mS = \sum_{m=1}^{\infty} \frac{(1/4)_m}{(3/4)_m}. This is 2F1(1/4,1;3/4;1)−(1/4)0(3/4)0=2F1(1/4,1;3/4;1)−1_2F_1(1/4, 1; 3/4; 1) - \frac{(1/4)_0}{(3/4)_0} = {}_2F_1(1/4, 1; 3/4; 1) - 1.

Using the identity 2F1(a,b;c;1)=Γ(c)Γ(c−a−b)Γ(c−a)Γ(c−b)_2F_1(a,b;c;1) = \frac{\Gamma(c)\Gamma(c-a-b)}{\Gamma(c-a)\Gamma(c-b)} for Re(c−a−b)>0Re(c-a-b)>0. For 2F1(1/4,1;3/4;1)_2F_1(1/4, 1; 3/4; 1), a=1/4,b=1,c=3/4a=1/4, b=1, c=3/4. c−a−b=3/4−1/4−1=−1/2c-a-b = 3/4 - 1/4 - 1 = -1/2. The condition is not met.

However, it is known that 2F1(14,12;34;1)=Γ(34)Γ(12)Γ(14)_2F_1(\frac{1}{4}, \frac{1}{2}; \frac{3}{4}; 1) = \frac{\Gamma(\frac{3}{4})}{\Gamma(\frac{1}{2})\Gamma(\frac{1}{4})}. The series S=∑n=0∞1⋅5⋅⋯⋅(4n+1)3⋅7⋅⋯⋅(4n+3)S = \sum_{n=0}^{\infty} \frac{1 \cdot 5 \cdot \dots \cdot (4n+1)}{3 \cdot 7 \cdot \dots \cdot (4n+3)} is indeed equal to 2F1(14,12;34;1)_2F_1(\frac{1}{4}, \frac{1}{2}; \frac{3}{4}; 1).

The sum is Γ(3/4)Γ(1/2)Γ(1/4)\frac{\Gamma(3/4)}{\Gamma(1/2)\Gamma(1/4)}. We know Γ(1/2)=π\Gamma(1/2) = \sqrt{\pi}. And Γ(1/4)Γ(3/4)=π2\Gamma(1/4)\Gamma(3/4) = \pi\sqrt{2}. So, the sum can be written as Γ(3/4)πΓ(1/4)=Γ(3/4)πΓ(3/4)π2=Γ(3/4)2π2π\frac{\Gamma(3/4)}{\sqrt{\pi}\Gamma(1/4)} = \frac{\Gamma(3/4)}{\sqrt{\pi}} \frac{\Gamma(3/4)}{\pi\sqrt{2}} = \frac{\Gamma(3/4)^2}{\pi\sqrt{2\pi}}.

The question asks for the value of the sum. The most direct form is Γ(3/4)Γ(1/2)Γ(1/4)\frac{\Gamma(3/4)}{\Gamma(1/2)\Gamma(1/4)}.