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?

$$\sum\limits_{n=0}^{+\infty} \frac{n+1}{(2n+1)!} = \frac{1}{2} \cdot \sum\limits_{n=0}^{+\infty} \frac{(2n+1)+1}{(2n+1)!} = \frac{1}{2} \cdot \sum\limits_{n=0}^{+\infty} \bigg(\frac{1}{(2n)!}+\frac{1}{(2n+1)!}\bigg) = \frac{1}{2} \sum\limits_{k=0}^{+\infty} \frac{1}{k!} = \frac{e}{2}$$

Maybe the details are not explained well enough. If there is anything unclear just ask.

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.