Evaluate $\frac {1+\frac {2^2}{2!} +\frac {2^4}{3!}+\frac {2^6}{4!} +\dots}{1+\frac {1}{2!}+\frac {2}{3!}+\frac {2^2}{4!}+\dots}$

Question

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.

Answer 1

$${1+\dfrac {1}{2!}+\dfrac {2}{3!}+\dfrac {2^2}{4!}+....} = \frac{1}{2^2}\bigg\{4+\dfrac {2^2}{2!}+\dfrac {2^3}{3!}+\dfrac {2^4}{4!}+....\bigg\}$$

$$=\frac{1}{2^2}\bigg\{1+1+2+\dfrac {2^2}{2!}+\dfrac {2^3}{3!}+\dfrac {2^4}{4!}+....\bigg\} = \frac{1}{4}\bigg(1+e^2\bigg)$$

Answer 2

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 $${e^2+1}$$