One of the best-known series is the geometric series. It is called by such a name because its general term is a geometric sequence. Here on this page, we give more details about the convergence of this series. This series has several applications in mathematical analysis and the probability theorem. There is a probability law called the geometric distribution, which is useful in statistics.

## Sum of geometric series proof

We define and discuss the convergence of the geometric series.

Definition: Let $r$ be a real number. The series $$\sum_{n=0}^{+\infty} r^n$$ is called the geometric series.

We will see below that the convergence of the above series depends on the position of the real number $r$.

If we set $u_n=r^n,$ then we have $$\frac{u_{n+1}}{u_n}=r.$$ Because of this, the number $r$ is called the common ratio of the geometric series.

Let $S_n=u_0+\cdots+u_n$ the nth partial sum the above series. If $r=1$, then $S_n=n\to+\infty$. Thus the series is divergent.

Now we assume that $r\neq 1$. By using the remarkable identities, we have $$S_n=\frac{1-r^{n+1}}{1-r}.$$

• If $r\in (-1,1)$. Then, according to the course of the geometric sequence, we have $r^{n+1}\to 0$ as $n\to \infty$. Thus the geometric series converges and $$\sum_{n=0}^{+\infty} r^n=\frac{1}{1-r}.$$
• if $r\ge 1$, then for any $n,$ $r^n\ge 1,$ so that $S_n=\sum_{k=1}^n r^k\ge \sum_{k=1}^n 1=n$. Thus $S_n\to +\infty$ as $n\to +\infty$. Hence the series diverges for any $r\ge 1,$ and $$\sum_{n=0}^{+\infty} r^n=+\infty.$$
• if $r\le -1$ then we have $$S_n=\frac{1+(-1)^{n+1} (-r)^n}{1-r}.$$ Thus the limit of this sequence does not exist. So the series divergent for any $r\le -1$.

### Applications to the convergence of series

Exercise: Prove the following series is convergent $$\sum_{n=0}^\infty \sin(e^{-n}).$$ Proof: We recall that the Euler number $e>1$. So that $0<e^{-1}<n$. This implies that the series $\sum_{n=0}^{+\infty} e^{-n}$ is convergent, due to the previous paragraph. Using the property $|\sin(x)|\le |x|$ for any $x\in\mathbb{R},$ we have $$|\sin(e^{-n})|\le e^{-n},\quad n\ge 0.$$ By using the comparaison test for series of positive terms, we deduce that the series $\sum_{n=0}^\infty \sin(e^{-n})$ is absolutely convergent, then it is a convergent series.

You may also consult the convergence of series using the ratio test for more examples and exercises with detailed answers.