<s>[INST] How to calculate the sum of the infinite series Please help me with the sum of the infinite series:
$$
   \Large \frac{1+\frac{\pi^4}{5!}+\frac{\pi^8}{9!}+\frac{\pi^{12}}{13!}+...}
   {\frac{1}{3!}+\frac{\pi^4}{7!}+\frac{\pi^8}{11!}+\frac{\pi^{12}}{15!}+...}
$$
 [/INST] 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:
$$\boxed{\pi^2}$$
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