Why the limit can only be $0$

Question

In my book it says if the limit $\lim\limits_{n \to \infty}\prod\limits_{i=1}^n(1+\varepsilon_it)$exsist ,it can only be $0$where $0<t<1,\varepsilon_i=1\ \mbox{or}\ -1$.

Why it can only be zero

Answer 1

Suppose the limit exists, say $L$, and it is not $0$. Then

$$1= \frac{L}{L} = \lim_{n \to \infty} \frac{\prod_{i=1}^{n+1} (1+ \varepsilon_it)}{\prod_{i=1}^{n} (1+ \varepsilon_it)} = \lim_{n \to \infty} 1+ \varepsilon_nt$$

And this is a contradiction, because the latter limit can be only $1+t >1$ (if $\varepsilon_n = 1$ eventually) or $1-t<1$ (if $\varepsilon_n = -1$ eventually), or it does not exist (otherwise).

Hence the limit must be $0$.

In general, using this argument, you can prove that if an infinite product converges to a nonzero limit $L$, then the terms of the product must converge to $1$. This is similar to the argument showing that in a converging series the terms must converge to $0$.