1. ## Quotient

This shouldn't be to heavy Challenge. Calculate

a) $
\frac{1+\frac{{\pi}^4}{5!}+\frac{{\pi}^8}{9!}+\fra c{{\pi}^{12}}{13!}+\cdots}{\frac 1{3!}+\frac{{\pi}^4}{7!}+\frac{{\pi}^8}{11!}+\frac {{\pi}^{12}}{15!}+\cdots}
$
(very easy)

b) $
\frac{1+\frac{{\pi}^4}{2^4\cdot 4!}+\frac{{\pi}^8}{2^8\cdot 8!}+\frac{{\pi}^{12}}{2^{12}\cdot 12!}+\cdots}{\frac 1{2!}+\frac{{\pi}^4}{2^4\cdot 6!}+\frac{{\pi}^8}{2^8\cdot 10!}+\frac{{\pi}^{12}}{2^{12}\cdot 14!}+\cdots}
$
(not so easy)

2. Originally Posted by ns1954
This shouldn't be to heavy Challenge. Calculate

a) $
\frac{1+\frac{{\pi}^4}{5!}+\frac{{\pi}^8}{9!}+\fra c{{\pi}^{12}}{13!}+\cdots}{\cdots}
$
(very easy)

b) $
\frac{1+\frac{{\pi}^4}{2^4\cdot 4!}+\frac{{\pi}^8}{2^8\cdot 8!}+\frac{{\pi}^{12}}{2^{12}\cdot 12!}+\cdots}{\frac 1{2!}+\frac{{\pi}^4}{2^4\cdot 6!}+\frac{{\pi}^8}{2^8\cdot 10!}+\frac{{\pi}^{12}}{2^{12}\cdot 14!}+\cdots}
$
(not so easy)

$\sin(x) = x - \frac{x^3}{3!} + \frac{x^5}{5!} - \frac{x^7}{7!} + ...$

Sub $x = \pi$ we obtain

$\sin(\pi ) = \pi - \frac{\pi ^3}{3!} + \frac{\pi ^5}{5!} - \frac{\pi ^7}{7!} + ... = 0$

rearrange the series ,

$\pi ( 1+\frac{{\pi}^4}{5!}+\frac{{\pi}^8}{9!}+\frac{{\pi }^{12}}{13!}+ ... ) = \pi^3 (\frac 1{3!}+\frac{{\pi}^4}{7!}+\frac{{\pi}^8}{11!}+\frac {{\pi}^{12}}{15!}+ ...)
$

so the answer to the first problem is $\pi^2$

3. You absolytly right about a). Try b) by using power series. Define apropriate function and use standard power series for some functions...

4. The second one is exactly like the first one, but using the series expansion of $\cos{x}$ instead of $\sin{x}$:

We have $\cos{x} = \sum_{n=0}^\infty (-1)^n \frac{x^{2n}}{(2n)!} =\sum_{n=0}^\infty \frac{x^{4n}}{(4n)!} - \sum_{n=0}^\infty \frac{x^{4n+2}}{(4n+2)!}$

Put $x=\frac{\pi}{2}$ to get $\sum_{n=0}^\infty \frac{\left(\frac{\pi}{2}\right)^{4n}}{(4n)!} - \sum_{n=0}^\infty \frac{\left(\frac{\pi}{2}\right)^{4n+2}}{(4n+2)!} = 0$ or equivalently $\sum_{n=0}^\infty \frac{\pi^{4n}}{2^{4n}(4n)!} = \frac{\pi^2}{4} \sum_{n=0}^\infty \frac{\pi^{4n}}{2^{4n}(4n+2)!}$.
Therefore:

$
\frac{\pi^2}{4} = \frac{\sum_{n=0}^\infty \frac{\pi^{4n}}{2^{4n}(4n)!}}{\sum_{n=0}^\infty \frac{\pi^{4n}}{2^{4n}(4n+2)!}} = \frac{1+\frac{{\pi}^4}{2^4\cdot 4!}+\frac{{\pi}^8}{2^8\cdot 8!}+\frac{{\pi}^{12}}{2^{12}\cdot 12!}+\cdots}{\frac 1{2!}+\frac{{\pi}^4}{2^4\cdot 6!}+\frac{{\pi}^8}{2^8\cdot 10!}+\frac{{\pi}^{12}}{2^{12}\cdot 14!}+\cdots}
$