Proving ratio of subsequent terms in sequence <1, then limit of sequence tends to 0 If $\lim_{n \rightarrow \infty} \left|\frac{u_{n+1}}{u_n}\right|=|a|<1$, prove that $\lim_{n \rightarrow \infty} u_n=0$.
We have to prove that for $\forall \epsilon>0: |u_n|<\epsilon, \forall n>N\in\mathbb{N}$
I started by:
$$\lim_{n \rightarrow \infty} \left|\frac{u_{n+1}}{u_n}\right|<1\implies 0<|u_{n+1}|<|u_n|$$
So $|u_n|$ is strictly monotonic decreasing and bounded below, so it converges to a value $u^*$. So, $\forall\epsilon<0:|u_n-u^*|<\frac{\epsilon}{2}, \forall n>N$. Hence, we can create a set 
$$I=\bigcap_{n\in\mathbb{N}}I_n=\bigcap_{n\in\mathbb{N}}[0,|u_n|]$$
Where $$I_n\subset I_{n+1}\subset I_{n+2}\subset...$$
Then we have to show there exists only one value in the set, namely $0$, so that $u^*=0$. Then for $n>N$:
$$u^*\leq |u^*-u_N|+|u_N-0|<\frac{\epsilon}{2}+\frac{\epsilon}{2}=\epsilon\implies u^*<\epsilon\implies u^*=0$$
Is this proof correct? I think I made a wrong assumption at the end, but I wouldn't know how else to do it.

I don't think you can say $|u_N - 0| \leq \frac{\epsilon}{2}$ without implicitly assuming $u^* = 0$. Here's what I would do.
Start from the point where you have defined $u^*$. Let $a < 1$ be the limit of the ratio of terms of the sequence. Then we have that $$\lim_{n \rightarrow \infty} \left| \frac{a_{n+1}}{a_n} \right| = a \\ \implies \\ \lim_{n \rightarrow \infty} |a_{n+1}| = \lim_{n \rightarrow \infty} |a_{n}| \left| \frac{a_{n+1}}{a_n} \right| = a \lim_{n \rightarrow \infty} |a_n| \\ \implies \\ u^* = au^* \\ \implies \\ u^*(1-a) = 0$$
Then as $a < 1$, $1 - a \not = 0$. So $u^* = 0$.
Alternatively, you can use the fact $$\limsup_{n \rightarrow \infty} \left| \frac{a_{n+1}}{a_n} \right| < 1 \implies \sum_{n \in \mathbb{N}} a_n \text{ converges} \implies a_n \rightarrow 0$$ if these results have already been proven in whatever setting you are working in.
