Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

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.

share|cite|improve this question
One thing i just realized while revisiting my notes which I missed was i have n't given any thought to ratio test. – Comic Book Guy 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 ? – Comic Book Guy 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 Gauss's critieron seems the same as the result i was stating although i did n't know it was called that. 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. – Comic Book Guy Mar 9 '12 at 23:29
up vote 14 down vote accepted

We can prove by induction that

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

and so your series diverges.

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

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

share|cite|improve this answer
For a proof approach, see this answer: 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.

share|cite|improve this answer
Sorry Buddy i could only pick one answer but i found your answer very slick and educational too. – Comic Book Guy Mar 9 '12 at 22:44

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.