How to calculate the sum of the infinite series Please help me with the sum of the infinite series:
$$
   \Large \frac{1+\frac{\pi^4}{5!}+\frac{\pi^8}{9!}+\frac{\pi^{12}}{13!}+...}
   {\frac{1}{3!}+\frac{\pi^4}{7!}+\frac{\pi^8}{11!}+\frac{\pi^{12}}{15!}+...}
$$
 A: The series for $\sinh{x}$ is
$$\sinh{x} = \sum_{k=0}^{\infty} \frac{x^{2 k+1}}{(2 k+1)!} $$
The fraction is 
$$\pi^2 \frac{g''(\pi)}{g(\pi)} $$
where
$$g(x) = \sum_{k=0}^{\infty} \frac{x^{4 k+1}}{(4 k+1)!} $$
We find $g$ by solving the equation
$$g''(x) + g(x) = \sinh{x}$$
$$g(0)=g'(0)=0$$
You may solve this equation via a Laplace transform, for example.  The solution is
$$g(x) = \frac12 \sinh{x} - \frac12 \sin{x}$$
$$g''(x) = \frac12 \sinh{x} + \frac12 \sin{x}$$
Therefore, the fraction is $\pi^2$.
A: Evaluating the numerator first.
Consider the series 
$$\sum_{n=0}^{\infty} \frac{x^{4n}}{(4n)!}=1+\frac{x^4}{4!}+\frac{x^8}{8!}+\cdots$$
Also,
$$\cos x=1-\frac{x^2}{2!}+\frac{x^4}{4!}-\frac{x^6}{6!}+\frac{x^8}{8!}-\cdots$$
$$\cosh x=\frac{x^2}{2!}+\frac{x^4}{4!}+\frac{x^6}{6!}+\frac{x^8}{8!}+\cdots$$
Add the above two and you obtain:
$$\sum_{n=0}^{\infty} \frac{x^{4n}}{(4n)!}=\frac{1}{2}(\cos x+\cosh x)\,\,\,\,(*)$$
Integrate both the sides from the limits $0$ to $\pi$ to get:
$$\sum_{n=0}^{\infty} \frac{\pi^{4n+1}}{(4n+1)!}=\frac{\sinh(\pi)}{2} \Rightarrow  \sum_{n=0}^{\infty} \frac{\pi^{4n}}{(4n+1)!}=\frac{\sinh(\pi)}{2\pi}\,\,\,\,\,\,(1)$$
For denominator, integrate both the sides of $(*)$ wrt $x$ thrice and substitute $x=\pi$ to obtain:
$$\sum_{n=0}^{\infty} \frac{\pi^{4n+3}}{(4n+3)!}=\frac{1}{2}\sinh(\pi)\Rightarrow \sum_{n=0}^{\infty} \frac{\pi^{4n}}{(4n+3)!}=\frac{\sinh(\pi)}{2\pi^3}\,\,\,\,(2)$$
Divide $(1)$ and $(2)$ to obtain:
$$\boxed{\pi^2}$$
A: Denominator:
You are certainly familiar with the series
$$\sin x=\frac{x^1}{1!}-\frac{x^3}{3!}+\frac{x^5}{5!}-\frac{x^7}{7!}+\ldots$$
What happens if we add to it the series for $i\sin(ix)=-\sinh x$?
Can you guess what would be the corresponding trick for the numerator?
A: Hint:
$$\sum \frac{x^{4k}}{(4k+1)!} = \frac{\sinh(x)+\sin(x)}{2x}$$
$$\sum \frac{x^{4k}}{(4k+3)!} = \frac{\sinh(x)-\sin(x)}{2x^3}$$
