<s>[INST] Evaluate $\frac {1+\frac {2^2}{2!} +\frac {2^4}{3!}+\frac {2^6}{4!} +\dots}{1+\frac {1}{2!}+\frac {2}{3!}+\frac {2^2}{4!}+\dots}$ Evaluate the given series 
$$\dfrac {1+\dfrac {2^2}{2!} +\dfrac {2^4}{3!}+\dfrac {2^6}{4!} +....}{1+\dfrac {1}{2!}+\dfrac {2}{3!}+\dfrac {2^2}{4!}+....}$$
If we factor out $\dfrac {1}{2^2}$ from the numerator we are left with 
$$\dfrac {2^2}{1!}+\dfrac  {(2^2)^{2}}{2!} + \dfrac {(2^2)^{3}}{3!}+.....$$ which is equal to $(e^2)^{2} -1$. But I couldn't manipulate the denominator. 
 [/INST] $${1+\dfrac {1}{2!}+\dfrac {2}{3!}+\dfrac {2^2}{4!}+....} = \frac{1}{2^2}\bigg\{4+\dfrac {2^2}{2!}+\dfrac {2^3}{3!}+\dfrac {2^4}{4!}+....\bigg\}$$ 
$$=\frac{1}{2^2}\bigg\{1+1+2+\dfrac {2^2}{2!}+\dfrac {2^3}{3!}+\dfrac {2^4}{4!}+....\bigg\} = \frac{1}{4}\bigg(1+e^2\bigg)$$
</s>