Find a formula for $\prod_{i=1}^{2n-1} \left(1-\frac{(-1)^i}{i}\right)$ then prove it.
I assumed that $\prod_{i=1}^{2n-1} \left(1-\frac{(-1)^i}{i}\right)=\frac{2n}{2n-1}$ after doing a few cases from above then I tried to prove it with induction would this be a fair approach or any other approaches that would work?
Induction is fine, but there is an easier way. Note that if $i$ is an odd integer then the product of $i$th factor and the next one is $$\left(1-\frac{(-1)^i}{i}\right)\left(1-\frac{(-1)^{i+1}}{i+1}\right)=\left(1+\frac{1}{i}\right)\left(1-\frac{1}{i+1}\right)=\frac{i+1}{i}\cdot \frac{i}{i+1}=1.$$ Hence, for an even number of factors, we get $$\prod_{i=1}^{2n-2} \left(1-\frac{(-1)^i}{i}\right)=1.$$ What may we conclude about the product $$\prod_{i=1}^{2n-1} \left(1-\frac{(-1)^i}{i}\right)\quad ?$$
It's not too hard to derive the solution directly:
$\Pi_{i=1}^{2n-1} (1-\frac{(-1)^i}{i})= \frac{1}{(2n-1)!}\Pi_{i=1}^{2n-1} (i-(-1)^i) $
$= \frac{1}{(2n-1)!}(1 +1)(2 - 1)(3+1)(4-1)\ldots(2n-1+1)$
$=\frac{1}{(2n-1)!}(2)(1)(4)(3)\ldots(2n-2)(2n-3)(2n)$
(note the lack of $(2n-1)$ term here)
$=\frac{1}{(2n-1)!}\frac{(2n)!}{(2n-1)}$
$=\frac{2n}{(2n-1)}$
