Prove that $1 + \frac{2}{3!} + \frac{3}{5!} + \frac{4}{7!} + \dotsb = \frac{e}{2}$ 
Prove that $1 + \frac{2}{3!} + \frac{3}{5!} + \frac{4}{7!} + \dotsb = \frac{e}{2}$.

This is problem 4 from page 303 of S.L.Loney's 'Plane Trigonometry'. 
It seems fairly obvious that the series expansion $e^x$ will be used. However, I am unsure where to start. Should I consider other series? 

$$e=(1+1)+\left(\frac {1}{2!}+\frac {1}{3!}\right) +\left(\frac {1}{4!}+\frac {1}{5!}\right) +\left(\frac {1}{6!}+\frac {1}{7!}\right) +\cdots=2\left(1+\frac {2}{3!}+\frac {3}{5!}+\frac {4}{7!}+\cdots\right) $$
Q. E. D
