# Infinite series : $\sum_{n=1}^{\infty}\prod_{i=1}^n\frac{(3i-1)}{(4i-3)}$

How to evaluate this? $$\sum_{n=1}^{\infty}\prod_{i=1}^n\frac{(3i-1)}{(4i-3)}$$

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Do you mean $$\sum_{n=1}^{\infty}\prod_{i=1}^n\frac{(3i-1)}{(4i-3)}$$ – Ross Millikan Feb 14 '13 at 13:28
The sum converges by the Ratio Test. Do you want the value of the sum? – David Mitra Feb 14 '13 at 13:31
@RossMillikan, Yes – gauss115 Feb 14 '13 at 13:44

You can sum this in terms of the hypergeometric function. Here is a result by maple

$$2\,{ _2F_1\left( 1, \frac{5}{3}; \,\frac{5}{4};\,\frac{3}{4}\right)}.$$

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But of course it is essentially the definition of that hypergeometric. – GEdgar Feb 14 '13 at 14:03
@GEdgar: It is the most possible closed form solution. – Mhenni Benghorbal Feb 28 '13 at 18:31

Nope it doesn't diverge the limit is $2 \cdot \text{Hypergeometric2F1}[1,\frac{5}{3},\frac{5}{4},\frac{3}{4}]$ (according to Mathematica) I Interpret it as $\sum_{n=1}^\infty \prod_{i=1}^n \frac{3i-1}{4i-3}$

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The Value is about 13.2047328961513303884487902215 – Dominic Michaelis Feb 14 '13 at 13:49

An answer is $$3^{-1/4}2^{7/3}B\left(\frac{1}{4},\frac{17}{12},\frac{3}{4}\right) \approx 13.2047328961513303884487902215$$ where we have used the incomplete Beta function $$B(\nu,\mu,x) := \int_0^x t^{\nu-1}(1-t)^{\mu-1}\;dt$$

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I took the hypergeometric and looked it up an a table
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