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Sum the series $$\frac{1^2}{2^2}+\frac{1^2\cdot3^2}{2^2\cdot4^2}+\frac{1^2\cdot3^2\cdot5^2}{2^2\cdot4^2\cdot6^2}+\ldots$$

Someone please help me in finding sum of this infinite series. I have never found the sum of this kind of series.

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    Thanks for editing.. can you give a hint how to solve this one – 2017-01-03
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    I think this is related to Wallis' formula. – 2017-01-03
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    To me, this looks like an opportunity to use a Taylor series - this series looks like a Taylor series evaluated at $x = 1$. Unfortunately, I'm not sure *which* Taylor series - I'll think about it and get back to you if I come up with anything, but maybe there's a series you've seen in class that would work. – 2017-01-03
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    According to this comment http://math.stackexchange.com/q/118388 it seems that the series is divergent by comparison with the harmonic series. – 2017-01-03
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    The series diverges. – 2017-01-03
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    Oh yes thanks..This series is divergent. – 2017-01-03
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    One may prove that $$\frac{1^2}{2^2}\cdot\frac{1}{4^2}+\frac{1^2\cdot3^2}{2^2\cdot4^2}\cdot\frac{1}{6^2}+\cdots+\frac{1^2\cdots(2n-1)^2}{2^2\cdot4^2\cdots(2n)^2}\cdot\frac{1}{(2n+2)^2}+\cdots=\frac4\pi-\frac54. $$ – 2017-01-03
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    @OlivierOloa - What an interesting observation! Could you please post the derivation somewhere? – 2017-01-03
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    @hypergeometric Thanks. The proof is a bit long to write down ;) One may start with Wallis integrals ending up with the evaluation of two elliptic integrals: https://en.wikipedia.org/wiki/Elliptic_integral#Incomplete_elliptic_integral_of_the_first_kind – 2017-01-03

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The function $${}_2 F_1(1/2,1/2;1;z) = 1 + \sum_{n=1}^{\infty} \frac{((1/2)(3/2)...(1/2 + n - 1))^2}{n!^2} z^{n} = 1 + \sum_{n=1}^{\infty} \Big( \frac{1 \cdot 3 \cdot ... \cdot (2n-1)}{2 \cdot 4 \cdot ... \cdot (2n)} \Big)^2 z^{n}$$

can be expressed as an elliptic integral $$\frac{2}{\pi} \int_0^{\pi/2} \frac{1}{\sqrt{1 - z \sin^2(\theta)}} \, \mathrm{d}\theta$$ whenever $|z| < 1.$

From the integral representation it is easy to see that diverges as $z$ approaches $1$. For fun, the alternating series with $z = -1$ can be evaluated as $$1 -\frac{1^2}{2^2} + \frac{1^2 \cdot 3^2}{2^2 \cdot 4^2} \pm ... = \frac{\Gamma(1/4)^2}{(2\pi)^{3/2}} \approx 0.8346$$

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One may observe that the general term of this series rewrites $$ a_n:=\frac{1^2\cdots(2n-1)^2}{2^2\cdot4^2\cdots(2n)^2}=\frac{[(2n)!]^2}{2^{4n}\cdot(n!)^4} $$ thus, as $n \to \infty$, it satisfies $$ \left|\frac{a_n}{a_{n+1}}\right| = 1+ \frac{1}{n} + \frac{C_n}{n^2},\quad |C_n|<1, $$ and the series diverges by Gauss' test which is a refined ratio test.

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    @Simple Art. Thanks! – 2017-01-03