Vtyjkyuk

Prove that $1 + frac23! + frac35! + frac47! + dotsb = frace2$.

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?

$$e=(1+1)+left(frac 12!+frac 13!right) +left(frac 14!+frac 15!right) +left(frac 16!+frac 17!right) +cdots=2left(1+frac 23!+frac 35!+frac 47!+cdotsright) $$

Q. E. D

$$sumlimits_n=0^+infty fracn+1(2n+1)! = frac12 cdot sumlimits_n=0^+infty frac(2n+1)+1(2n+1)! = frac12 cdot sumlimits_n=0^+infty bigg(frac1(2n)!+frac1(2n+1)!bigg) = frac12 sumlimits_k=0^+infty frac1k! = frace2$$

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

Another approach is
$$sinh x=x+dfracx^33!+dfracx^55!+dfracx^77!+cdots$$
then
$$(xsinh x)'Big|_x=1=2left(1+dfrac23!+dfrac35!+dfrac47!+cdotsright)$$
which gives the result.

Your series is $$sum_n=1^inftyfracn(2n-1)!.$$
Let $a_n=fracn(2n-1)!$, then
$$a_n=frac12frac2n(2n-1)!=frac12frac2n-1+1(2n-1)!
=frac12left(frac1(2n-2)!+frac1(2n-1)!right).$$

So $$sum_n=1^inftyfracn(2n-1)!=frac12sum_n=1^inftyleft(frac1(2n-2)!+frac1(2n-1)!right)$$
$$=frac12sum_n=0frac1n!=frac12e.$$