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. 

Series in numerator is the expansion of $e^x-1$ at $x=4$ and series denomirator is the expansion of $e^x-1$ at $x=2$. Substituting the values we get $$\frac{e^4-1}{e^2-1}$$
which on simplification gives $${e^2+1}$$
