<s>[INST] 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? 
 [/INST] Your series is $$\sum_{n=1}^{\infty}\frac{n}{(2n-1)!}.$$
Let $a_n=\frac{n}{(2n-1)!}$, then 
$$a_n=\frac{1}{2}\frac{2n}{(2n-1)!}=\frac{1}{2}\frac{2n-1+1}{(2n-1)!}
=\frac{1}{2}\left(\frac{1}{(2n-2)!}+\frac{1}{(2n-1)!}\right).$$
So $$\sum_{n=1}^{\infty}\frac{n}{(2n-1)!}=\frac{1}{2}\sum_{n=1}^{\infty}\left(\frac{1}{(2n-2)!}+\frac{1}{(2n-1)!}\right)$$
$$=\frac{1}{2}\sum_{n=0}\frac{1}{n!}=\frac{1}{2}e.$$
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