If $\vert a \vert < 1$, then $\lim_{n\to \infty} a^n=0$ I'm trying to prove that if $\vert a \vert <1$, then $\lim_{n\to \infty} a^n=0$. Proving this with a contradiction did not lead anywhere, so I'm now approaching it via the binomial theorem. Lets write $a^n=(1+(a-1))^n$. Expanding this gives $$1^n + n(a-1) + ... + (a-1)^n$$
Now we have to look at two cases, i ) where $a>0$ and ii ) where $a<0$. i ) If $$a>0\rightarrow\vert a \vert=a$$ $$a<1 \to a-1<0$$
Lets look at $(a-1)$. Since $\vert a \vert <1$, this term takes values from the interval $(-1,0)$. This is where I'm stuck. It seems like we are reducing more and more from $1^n=1$, so naturally as $n \rightarrow \infty$ we have that $a^n \rightarrow 0$. How should I proceed from here?
 A: If $|a|<1$ then the inequality $|a|^n<\epsilon$ with $\epsilon>0$ is equivalent to $n\log(|a|)<\log(\epsilon)$, i.e.
$$n>\frac{\log(\epsilon)}{\log(|a|)}$$
where the inequality is now reversed because $\log(|a|)<0$.
Finally, apply the definition of limit.
A: First of all, split the cases by $|a|>1$ and $|a|<1$, then for the case of $|a|>1$, you can suppose it is bounded above and prove for contradiction. For the case of $|a|<1$, you can firstly prove that $a^{n}>0, \forall n \in \mathbb{N}$. Then since $a^{n+1}<a^{n}$, it forms a decreasing sequence, bounded below. By letting $L=\lim a^{n}$, you can try and prove that $L=0$, using $a^{n+1}=a \cdot a^{n}$.
A: First note that because of $|a^n| = |a|^n$ it suffices to consider the case $0 \le a < 1$. The case $a=0$ is trivial, so we are left with the case $0 < a < 1$.
Your idea of using the binomial formula now works if you apply it to the reciprocal $b = 1/a > 1$:
$$
 b^n = (1+(b-1))^n \ge 1 + n (b-1)
$$
and therefore
$$
 0 \le a^n = \frac{1}{b^n} \le \frac{1}{1+n(b-1)}
$$
and the right-hand side converges to $0$ for $n \to \infty$.
A: By Bernoulli's inequality, $$\frac1{(1+\epsilon)^n}<\frac1{1+n\epsilon}<\frac1{n\epsilon}.$$ 
This allows to get rid of the power.
A: We set $b:=|a|$ . Then we have $\lim_{n\to \infty} a^n=0 \iff \lim_{n\to \infty} b^n=0.$
Hence we show that $\lim_{n\to \infty} b^n=0.$ If $b=0$, we are done, hence let $b>0.$
Since $1/b >1$, there is $t>0$ such that $1/b=1+t$, thus $1/b^n=(1+t)^n \ge 1+nt$, by Bernoulli. This gives
$$b^n \le \frac{1}{1+nt} < \frac{1}{nt}.$$
Can you proceed ?
A: First consider the case where $a$ is positive
By the binomial theorem,
$(1+x)^n = 1+nx+. . . ≥ 1+nx$
now consider $(1+x)^{-n} \le \frac {1}{(1+nx)}$
set $ a = (1 + x)^{-1}$ since $a < 1$ then $x > 0$
therefore $a^n = (1+x)^{-n} \le \frac {1}{(1+nx)} \rightarrow 0$ as $n \rightarrow \infty$
The case where a is negative is done in the same way
