Is this an allowed step for working with infinite sequences? I just started learning about infinite sequences, which are of course very interesting. Out of curiosity, I tried doing a proof that:
$${1,-1,1,-1 ...} = 0$$
This was pretty easy to do, if I could make a certain step. Now, this step makes a lot of intuitive sense to me, but I would like to know whether this is actually mathematically justifiable. Does it work in every scenario? The step is:
$$ \lim_{n \to \infty} (-1)^n =   \lim_{n \to \infty} (-1)^{2n} + \lim_{n \to \infty} (-1)^{2n+1} $$
Or more generally:
$$ \lim_{n \to \infty} (a)^n =   \lim_{n \to\infty} (a)^{2n} + \lim_{n \to\infty} (a)^{2n+1} $$
The proof from there on isn't very difficult, so this is kind of the breaking point of my argument. Some insights on the matter would be very much appreciated!
 A: What you can say is that if $\lim_{n\to\infty}a_{2n}$ and $\lim_{n\to\infty}a_{2n+1}$ both exist and are equal, then $\lim_{n\to\infty}a_n$ exists and equals those limits.
Moreover, in the traditional sense of limits, $(-1)^n$ does not converge at all, it has no limit.
A: What you're trying to prove is not true.
If you have learned the formal definition of limit, it ought to be easily for you to prove directly from the definition that $0$ is not the limit of $(-1)^n$.
(Set $\varepsilon=\frac12$ and see that no possible $N$ can even begin to work).

In fact, your proposed rule
$$ \lim_{n \to \infty} a_n =   \lim_{n \to\infty} a_{2n} + \lim_{n \to \infty} a_{2n+1} $$
will yield clear falsehoods even for well-behaved sequences -- consider for example the sequence $1,1,1,1,\ldots$ whose limit is obviously $1$, so your rule would claim that $1=1+1$.
You may be confusing sequences for series: It is true that
$$ \sum_{n=0}^\infty a_n = \sum_{n=0}^\infty a_{2n} + \sum_{n=0}^\infty a_{2n+1} $$
if both series on the right-hand side converge. (On the other hand, if neither of the right-hand series converge, it is still possible that the one on the left will).
