Splitting infinite products up in a even and a odd part 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.
 A: This is not true for general infinite products.  If we define $a_{2n}=10^n$ and $a_{2n+1}= 10^{-n}$, then the infinite product
$$
\prod_{n=0}^\infty a_n
$$
is not defined, since the sequence of partial products is not convergent.  
In your special case, however, you can get the result you want in this way.  You'll need to be able to evaluate the finite product
$$
\prod_{k=2}^N1+\frac{(-1)^k}{k}\,,
$$
(hint: you will likely need a different expression for odd and even $k$ - in fact, for even $k$, this product should come out to be $1$), and then show that that sequence converges to $1$ as $N\to\infty$.
A: 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!
A: Given 
(1)$$f(2)f(3)f(4)f(5)...$$we can arrange the terms as 
(2)$$[f(2)f(3)][f(4)f(5)]...$$ If we define $g(2k)=f(k)f(k+1)$, then (2) becomes
(3)$$g(2)g(4)...$$ If we define $j=2k$, then (3) becomes
(4)$$\prod_{j=1}^{\infty}g(j)$$Since $k$ and $j$ are dummy variables, it doesn't matters what we use; $\prod_{j=1}^{\infty}g(j)$ is the same as $\prod_{k=1}^{\infty}g(k)$. In your case, $g(k) = 1$, so you conclude that the limit is 1.
There is, however, a complication. Going from (1) to (2) used the associative property of multiplication, which hold for finite products. One does, however, have to be careful about using it in infinite products. An example of whether the associative property doesn't hold is $(1-1)+(1-1)+(1-1)... \neq 1+(-1+1)+(-1+1)...$. That's with sums, but a similar problem occurs with products. Basically, the problem is that you're taking partial products after an even number of terms, which means that you are getting a subsequence of partial products. If the subsequence of partial products, and the entire series, both have limits, then those limits are equal. But the subsequence having a limit does not mean that the entire sequence has a limit. So to be rigorous, you would first have to show that your infinite product has a limit, then use your argument to say that the limit is 1.
