# convergence of $\sum \limits_{n=1}^{\infty }\bigl\{ \frac {1\cdot3 \cdots (2n-1)} {2\cdot 4\cdots (2n)}\cdot \frac {4n+3} {2n+2}\bigr\} ^{2}$

I am investigating the convergence of $$\begin{split}\sum _{n=1}^{\infty }\left\{ \dfrac {1\cdot 3\cdots (2n-1)} {2\cdot 4\cdots (2n)}\cdot \dfrac {4n+3} {2n+2}\right\} ^{2} &= \sum _{n=1}^{\infty }\left\{ \dfrac {\prod _{t=1}^n (2t-1)} {\prod _{t=1}^n (2t)}\cdot \dfrac {4n+3} {2n+2}\right\} ^{2} \\ &=\sum _{n=1}^{\infty }\left\{ \prod _{t=1}^n\left( 1-\dfrac {1} {2t}\right) \dfrac {4n+3} {2n+2}\right\} ^{2} \end{split}$$ which after some manipulations I have reduced to $$\sum _{n=1}^{\infty }e^ \left\{ 2\ln \left(2 -\dfrac {1} {2n+2}\right) +2\cdot \sum _{t=1}^{n}\ln \left( 1-\dfrac {1}{2t}\right) \right\}$$ and from an alternative approach I was able to reduce it to $$\sum _{n=1}^{\infty } \dfrac{\left( 4n+3\right) ^{2}}{4\left(n+1\right)^{2}} \prod _{t=1}^n\left( 2+\dfrac{1}{2t^{2}}-\dfrac{2}{t}\right)$$ I am unsure how to proceed from here in either of the two cases. Any help would be much appreciated.

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One thing i just realized while revisiting my notes which I missed was i have n't given any thought to ratio test. – Hardy Mar 9 '12 at 22:23
It seems to me that ratio test is inconclusive, because $\displaystyle \lim_{n\to \infty} \frac{a_{n+1}}{a_n} =1$. – Pacciu Mar 9 '12 at 22:28
Actually i recall a result if $\left| \dfrac {u_{n+1}} {u_{n}}\right| =1+\dfrac {A_{1}} {n }+O\left( \dfrac {1} {n^{2}}\right)$, where $A_{1}$ is independent of $n$, then the series is absolutely convergent if $A_{1} < -1$. Is that help full here ? – Hardy Mar 9 '12 at 22:34
YOu might want to take a look at Runge's or Gauss' critieron for convergence. – Pedro Tamaroff Mar 9 '12 at 23:22
@PeterT.off Gauss's critieron seems the same as the result i was stating although i did n't know it was called that. encyclopediaofmath.org/index.php/Gauss_criterion Do u have link for me to refer to Runge's criterion, brief google searches for that seems to bring up Runge-Kutta's criterion much more frequently. – Hardy Mar 9 '12 at 23:29

We can prove by induction that

$$\dfrac {1\cdot 3\cdots (2n-1)} {2\cdot 4\cdots (2n)} \ge \frac{1}{\sqrt{4n}}$$

You can also notice that

$$\dfrac {1\cdot 3\cdots (2n-1)} {2\cdot 4\cdots (2n)} = \dfrac{\binom{2n}{n}}{4^n}$$

and try using the approximation

$$\dfrac{\binom{2n}{n}}{4^n} = \frac{1}{\sqrt{\pi n}} \left(1 + \mathcal{O}\left(\frac{1}{n}\right)\right)$$

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For a proof approach, see this answer: math.stackexchange.com/a/9379/1102. It should be quite similar. – Aryabhata Mar 9 '12 at 22:34
a very powerful inequality (+1) – user 1618033 Aug 17 '12 at 19:15

Denote by $a_n$ the general term, which is positive. We can rewrite it as $\left(\frac{(2n)!}{4^nn!n!}\right)^2\left(\frac{4n+3}{2n+2}\right)^2$, which is equivalent to $b_n:=4\left(\frac{(2n)!}{4^nn!n!}\right)^2$. Now we use Stirling's formula, which states that $n!\overset{+\infty}{\sim}\left(\frac ne\right)^n\sqrt{2\pi n}$. We get \begin{align*} b_n&\overset{+\infty}{\sim} 4\left(\frac{\left(\frac{2n}e\right)^{2n}\sqrt{4n\pi}}{4^n\left(\frac ne\right)^{2n}2\pi n}\right)^2\\ &=\frac 4{n\pi}, \end{align*} and using the fact that the harmonic series diverges, we get that the series $\sum_n a_n$ is divergent.

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Sorry Buddy i could only pick one answer but i found your answer very slick and educational too. – Hardy Mar 9 '12 at 22:44