We provide you with all details about the geometric sequence. In fact, this simple sequence appears frequently in many subjects such as series and probability. Let us discover together this magic sequence.

## What is a geometric sequence?

Let $a$ be a real number, we denote $a\in\mathbb{R},$ where $\mathbb{R}$ is the set of real numbers. A sequence of the form $$u_n=a^n,\quad n\in\mathbb{N},$$ is called a geometric sequence.

Let us discuss the convergence of this sequence. This means looking at the behavior of $u_n$ when $n$ is very large. As this sequence depends on $a,$ logically the study of convergence also depends on the number $a$.

Let us start with the simple case $a=1$. Then $u_n=1$ for all $n$. This is the constant sequence equal to $1$. Thus it converges to the same value of $1$.

Assume that $a=-1$. Then $u_n=(-1)^n$. This sequence is not convergent, because $u_{2n}=1$ converges to $1,$ while $u_{2n+1}=-1$ converges to -1. We recall the following result: As sequence $(u_n)_n$ is convergent to a real number $\ell$ if and only if the sequences $(u_{2n})_n$ and $(u_{2n+1})_n$ converge to the same number $\ell$.

Now assume that $a\in (-1,1)$, this means that the absolute value of $a$ satisfies $|a|<1$. We will prove that in this case, we have $$\lim_{n\to\infty} u_n=0.$$ Remark that $|u_n|=|a^n|\le |a|^n$. Thus $-|a|^n<u_n<|a|^n.$ Now according to the squeeze theorem, to prove that the limit of the sequence $(u_n)$ is zero, it suffices to prove that $|a|^n$ goes to zero as $n$ goes to $+\infty$.

First, we need to be sure that the sequence $(|a|^n)_n$ is convergent. In fact, for all $n,$ we have $0\le |a|^n<1$. Moreover, the sequence $(|a|^n)_n$ is decreasing. This is because $|a|^{n+1}-|a|^n=(|a|-1) |a|^n\le 0$, as $|a|<1$. Thus the sequence $(|a|^n)_n$ is convergent, and thus, there exists a real number $\ell\in\mathbb{R}$ such that $\lim_{n\to\infty}|a|^n=\ell.$

Let’s prove that $\ell=0$. In fact, remark that $\lim_{n\to \infty}|a|^{n+1}=0$, think about this fact: when $n$ is large, $n+1$ is also large. Now the fact that $|a|^{n+1}-|a|^n=(|a|-1) |a|^n$ and letting $n\to\infty,$ we get $0=(|a|-1)\ell$. But $|a|\neq 1,$ this implies that $\ell=0$. This ends the proof.

Proof: By using the remarkable identities we have \begin{align*} v_n= \frac{1-\left(\frac{1}{3}\right)^{n+1}}{1-\frac{1}{3}}=\frac{3}{2}\left(1-\left(\frac{1}{3}\right)^{n+1}\right).\end{align*} According to the previous section, we have $$\lim_{n\to+\infty} \left(\frac{1}{3}\right)^{n+1}=0.$$ Thus $$\lim_{n\to+\infty}v_n=\frac{3}{2}.$$
Proof: We recall that $|\sin(x)|\le |x|$ for any $x\in \mathbb{R}$. Now we can estimate \begin{align*} |v_n|=\left|\sin\left(\frac{1}{2^n}\right) \right|\le \left(\frac{1}{2}\right)^n. \end{align*} This means that $$-\left(\frac{1}{2}\right)^n\le v_n\le \left(\frac{1}{2}\right)^n$$ for all $n$. It suffices to apply the squeeze theorem since by the previous section the geometric sequence $(\frac{1}{2^n})_n$ goes to zero as $n\to\infty$. Hence $$\lim_{n\to+\infty}v_n=0.$$