18
$\begingroup$

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!

$\endgroup$
1
  • $\begingroup$ Guys, thanks a lot for your nice answers (+1)! The value of the series is really beautiful. $\endgroup$ Commented Sep 9, 2012 at 5:51

3 Answers 3

27
$\begingroup$

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 $$ \sum_{n=1}^\infty\tfrac{1\cdot2\cdot3\cdots(2n-1)}{2\cdot4\cdot6\cdots2n}\frac{1}{2n+1}=\sin^{-1}(1)-1=\frac\pi2-1 $$

$\endgroup$
0
8
$\begingroup$

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.

$\endgroup$
8
$\begingroup$

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 $(396)$) !

$\endgroup$
1
  • $\begingroup$ In Mike Spivey's answer to my only question so far, the central binomial coefficient also appears in the denominator. $\endgroup$
    – robjohn
    Commented Sep 8, 2012 at 23:09

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .