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? 

Another approach is 
$$\sinh x=x+\dfrac{x^3}{3!}+\dfrac{x^5}{5!}+\dfrac{x^7}{7!}+\cdots$$
then
$$(x\sinh x)'\Big|_{x=1}=2\left(1+\dfrac{2}{3!}+\dfrac{3}{5!}+\dfrac{4}{7!}+\cdots\right)$$
which gives the result.
