While working on infinite products I was faced with the task to proof
$$\prod_{k=2}^{\infty}\left(1+\frac{(-1)^k}{k}\right)~=~1$$
which is quite easy by using the fact that a even and the following odd factor are cancelling each other out
$$\left(1+\frac{(-1)^{2k}}{2k}\right)\left(1+\frac{(-1)^{2k+1}}{2k+1}\right)~=~\left(\frac{2k+(1)^k}{2k}\right)\left(\frac{(2k+1)-(1)^k}{2k+1}\right)~=~\frac{2k+1}{2k}\frac{2k}{2k+1}$$
where the last term always equals $1$.
So I asked myself whether it is allowed to rewrite the infinite product in terms of even and odd factors or not. To point it out are these two definitions
$$\prod_{k=2}^{\infty}\left(\frac{k+(-1)^k}{k}\right)~=~\prod_{k=1}^{\infty}\frac{2k+1}{2k}\frac{2k}{2k+1}$$
equal? My intuition says yes hence both products are going from $k=0$ till infinity and so it should be irrelevant but I am not sure how to justify this intuition by myself.
Best Answer
Note that the product $\prod_{k=2}^{2K}\left(1+\frac{(-1)^k}{k}\right)$ can be written in terms of even and odd terms as
$$\begin{align} \prod_{k=2}^{2K}\left(1+\frac{(-1)^k}{k}\right)&=\prod_{k=1}^K \left(1+\frac{(-1)^{2k}}{2k}\right)\prod_{k=1}^{K-1} \left(1+\frac{(-1)^{2k+1}}{2k+1}\right)\\\\ &=\prod_{k=1}^K \left(\frac{2k+1}{2k}\right)\prod_{k=1}^{K-1} \left(\frac{2k}{2k+1}\right)\\\\ &=\left(1+\frac1{2K}\right)\prod_{k=1}^{K-1} \left(\frac{2k+1}{2k}\frac{2k}{2k+1}\right)\\\\ &=\left(1+\frac1{2K}\right)\tag1 \end{align}$$
while the product $\prod_{k=2}^{2K+1}\left(1+\frac{(-1)^k}{k}\right)$ can be written
$$\begin{align} \prod_{k=2}^{2K+1}\left(1+\frac{(-1)^k}{k}\right)&=\prod_{k=1}^K \left(1+\frac{(-1)^{2k}}{2k}\right)\prod_{k=1}^{K} \left(1+\frac{(-1)^{2k+1}}{2k+1}\right)\\\\ &=\prod_{k=1}^K \left(\frac{2k+1}{2k}\right)\prod_{k=1}^{K} \left(\frac{2k}{2k+1}\right)\\\\ &=\prod_{k=1}^{K} \left(\frac{2k+1}{2k}\frac{2k}{2k+1}\right)\\\\ &=1\tag2 \end{align}$$
Inasmuch as the right-hand sides of both $(1)$ and $(2)$ approach $1$ as $K\to\infty$, we conclude
$$\prod_{k=2}^{\infty}\left(1+\frac{(-1)^k}{k}\right)=1$$
as was to be shown!