Limit of a fraction at infinity when all terms go to zero Suppose $Y = \sum_{j = 1}^{(n-1)!} \prod_{i = 1}^{n-1} y_{ij}$ and $Z = \sum_{j = 1}^{(n-1)!} \prod_{i = 1}^{n-1} z_{ij}$. Also suppose $0 < x < 1$, $-1 < y_{ij} < 1$, and $-1 < z_{ij} < 1$ for all $i$ and $j$, and $x \geq |y_{ij}|$ and $x \geq |z_{ij}|$ for all $i$ and $j$, but it is such that each product has at least one $|y_{ij}| < x$ (and $|z_{ij}| < x$). I believe the following limit is true.
$
\lim_{n \rightarrow \infty}\frac{x^{n-1} + Y}{x^{n-1} + Y + Z} = 1
$
Loosely speaking, I believe the reason is because all other terms converge to zero faster than $x^{n-1}$, so we are left with
$
\lim_{n \rightarrow \infty}\frac{x^{n-1} + Y}{x^{n-1} + Y + Z} = \frac{x^{n-1} + 0}{x^{n-1} + 0 + 0} = 1
$.
How would one formally prove this?
Edit: Suppose, instead, all the $y_{ij}$ and $z_{ij}$ are weakly positive and $Y$ and $Z$ are such that $Y = \sum_{j = 1}^{n!}(-1)^j \prod_{i = 1}^{n} y_{ij}$ and $Z = \sum_{j = 1}^{n!} (-1)^j \prod_{i = 1}^{n} z_{ij}$.
Then is the above limit equal to $1$? If so, how could one show this?
 A: One short remark: Everything looks nicer in your above question if you replace $n-1$ by $n$; and since you only care about limits $n \to \infty$, that's okay to do, so I will do that in the following.
Your hopes do not work out, I think. One aspect that you seem to neglect is that your have a lot of summands in $Y$ and $Z$ because of the factorial. This can cancel out the smallness of the $y_{ij}$ and $z_{ij}$.
Assume $x = \tfrac{1}{2}, y_{ij} = 0$ and $z_{ij} = \tfrac{1}{4}$ for all $i,j$. These values fulfil your requirements. Then:
$$Y = 0, \quad Z = \tfrac{n!}{4^{n}}.$$
Then, we have
$$\frac{Z}{x^n} = \frac{n!}{4^n} \cdot 2^n = \frac{n!}{2^n} \stackrel{n \to \infty}{\to} \infty.$$
Hence
$$\frac{x^n + Y}{x^n + Y + Z} = \frac{1}{1 + \tfrac{Z}{x^n}} \stackrel{n \to \infty}{\to} \frac{1}{1 + \infty} = 0 \neq 1.$$
EDIT: For the edit of the question (the alternating sums), I will assume "weakly positive" means "greater or equal to zero". Consider $0 < x < 1$ arbitrary, $y_{ij} = 0$
$$z_{ij} = \begin{cases} \frac{x}{i+1} &\text{ if } j = 1, \\ x &\text{ else.}\end{cases}$$
These values again fulfil the requirements, and we have $Y = 0$ and
$$ Z = \sum_{i=1}^{n!} (-1)^i \prod_{j=1}^n z_{ij} = x^n \cdot \sum_{i=1}^{n!} (-1)^i \frac{1}{i+1}. $$
The sum at the end is related to the alternating harmonic series, and for $n \to \infty$ converges to $\log(2) - 1$.
Hence, again
$$\frac{x^n + Y}{x^n + Y + Z} = \frac{1}{1 + \tfrac{Z}{x^n}} \stackrel{n \to \infty}{\to} \frac{1}{1 + \sum_{i=1}^{\infty} (-1)^i \frac{1}{i+1}} = \frac{1}{\log(2)} \neq 1.$$
Alternatively, you could do something by setting $y_{ij} = z_{ij} = 0$ for all even $i$, since then the sum does not feel the alternation anymore. Then, in all the odd summands, you could do something like in my first example.
The problem seems to be that you do not impose any strong requirements on how the different $y_{ij}$ relate to one another. If they somehow all "uniformly" approached zero as you increase $i$, you may be able to make stronger statements, but just having a single $y_{ij}$ in every product be smaller than x is unfortunately not sufficient. In my first example, you even see that $Z$ does not converge to zero, it goes to infinity, because of the freedom you have when choosing the $z_{ij}$.
