Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

Compute the sum: $$\sum_{n=1}^{\infty}\dfrac {1\cdot 3\cdots (2n-1)} {2\cdot 4\cdots (2n)\cdot (2n+1)}$$ At the moment, I only know that it's convergent and this is not hard to see if you look at the answers here I received for other problem with a similar series. For the further steps I need
some hints if possible. Thanks!

share|improve this question
    
Guys, thanks a lot for your nice answers (+1)! The value of the series is really beautiful. –  Chris's sis Sep 9 '12 at 5:51

3 Answers 3

up vote 26 down vote accepted

Starting with the power series derived using the binomial theorem, $$ (1-x)^{-1/2}=1+\tfrac12x+\tfrac12\tfrac32x^2/2!+\tfrac12\tfrac32\tfrac52x^3/3!+\dots+\tfrac{1\cdot2\cdot3\cdots(2n-1)}{2\cdot4\cdot6\cdots2n}x^n+\cdots $$ and integrating, we get the series for $$ \sin^{-1}(x)=\int_0^x(1-t^2)^{-1/2}\mathrm{d}t=\sum_{n=0}^\infty\tfrac{1\cdot2\cdot3\cdots(2n-1)}{2\cdot4\cdot6\cdots2n}\frac{x^{2n+1}}{2n+1} $$ Setting $x=1$, we get $$ \sum_{n=1}^\infty\tfrac{1\cdot2\cdot3\cdots(2n-1)}{2\cdot4\cdot6\cdots2n}\frac{1}{2n+1}=\sin^{-1}(1)-1=\frac\pi2-1 $$

share|improve this answer

Let's rewrite this as : $$\sum_{n=1}^{\infty}\frac {1\cdot 3\cdots (2n-1)} {2\cdot 4\cdots (2n)\cdot (2n+1)}=\sum_{n=1}^{\infty}\frac {(2n)!} {4^n (n!)^2(2n+1)}$$ But this is merely the expansion of $\arcsin(x)-1$ for $x=1$ as found at this page on central binomial series : $$\arcsin(x)=2\sum_{n=0}^{\infty}\frac{\binom{2n}{n}}{2n+1}\left(\frac x2\right)^{2n+1}$$

That is $\ \displaystyle \arcsin(1)-1=\frac {\pi}2-1$

The previous link to Boris Gourévitch work concerning $\pi$ is rather interesting because of the many relations provided concerning $\binom{2n}{n}$ and because it shows that the central binomial term may be at the numerator as well as at the denominator (see $(378)$).
It will be at the numerator for $\arcsin(x)$ and at the denominator for $\arcsin(x)^2$ and $\dfrac{\arcsin(x)}{\sqrt{1-x^2}}$ (relation $(396)$) !

share|improve this answer
    
In Mike Spivey's answer to my only question so far, the central binomial coefficient also appears in the denominator. –  robjohn Sep 8 '12 at 23:09

We have $$\sum_{n=1}^{\infty}\dfrac {1\cdot 3\cdots (2n-1)} {2\cdot 4\cdots (2n)\cdot (2n+1)} = \sum_{n=1}^\infty {2n\choose n}\frac{1}{4^n(2n+1)}.$$ Now notice that $$\int_0^{1/2}dx\, 2x^{2n} = \frac{1}{4^n(2n+1)}.$$ Thus, we need only compute the sum $$\begin{eqnarray*} 2\sum_{n=1}^\infty {2n\choose n}x^{2n} &=& 2\sum_{n=0}^\infty {2n\choose n}x^{2n} - 2 \\ &=& \frac{2}{\sqrt{1-4x^2}} - 2 \end{eqnarray*}$$ and integrate.

share|improve this answer

Your Answer

 
discard

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.