show that the sum = sqrt(2)

**bgbgbgbg** · June 10th 2010, 09:08 AM

1+(1/4) + (1*3)/(4*8) + (1*3*5)/(4*8*12)+(1*3*5*7)/(4*8*12*16)+.......
=sqrt(2)

**Bruno J.** · June 10th 2010, 09:57 AM

Nevermind! Listen to tonio.

**tonio** · June 10th 2010, 10:27 AM

Originally Posted by bgbgbgbg

1+(1/4) + (1*3)/(4*8) + (1*3*5)/(4*8*12)+(1*3*5*7)/(4*8*12*16)+.......
=sqrt(2)

That's the sum $1+\frac{1}{4}+\sum^\infty_{n=2}\frac{1\cdot 3\cdot\ldots\cdots (2n-1)}{4\cdot (4\cdot 2)\cdot\ldots\cdot(4n)}$ . Let's try now to understand the quotient in that infinite series a little better:

$\frac{1\cdot 3\cdot\ldots\cdot (2n-1)}{4\cdot (4\cdot 2)\cdot\ldots\cdot (4n)}=\frac{1\cdot 2\cdot 3\cdot 4\cdot\ldots\cdot (2n)}{[4^n\cdot n!][2\cdot 4\cdot\ldots\cdot (2n)]}$ $=\frac{(2n)!}{n!4^n\cdot n! 2^n}$ $=\frac{(2n)!}{(n!)^22^{3n}}$ $=\frac{(2n)!}{2^{2n}(n!)^2}\,\left(\frac{1}{\sqrt{ 2}}\right)^{2n}$ . There's a reason why we wrote the last mess:

We know (hopefully!) that $\arcsin x=\sum^\infty_{n=0}\frac{(2n)!}{2^n(n!)^2(2n+1)}\, x^{2n+1}$ , with convergence radius $|x|<1$ . Well, derivate both sides of this and get:

$\frac{1}{\sqrt{1-x^2}}=\sum^\infty_{n=0}\frac{(2n)!}{2^{2n}(n!)^2}\ ,x^{2n}$ , and now input $x=\frac{1}{\sqrt{2}}$ , do a little algebra and get your result.

Tonio

Ps. This is, imho, a problem that fits the Challenge Problems Section!

**awkward** · June 10th 2010, 02:30 PM

Another way to view the series is that it is the expansion by the Binomial Theorem of

$\left(1 - \frac{1}{2} \right)^{-1/2}$ .

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