Séries infinitas geometricas

When it comes to infinite series, there are several very common types. Perhaps the most frequently occurring infinite series is a Geometric Series. Before discussing Geometric Series let's review Infinite Series . Here are some things to keep in mind about infinite series: $\\$ $$\bullet\text{ every infinite series } \sum_{k=1}^{\infty} a_k \text { has a sequence of partial sums } \\ (S_1=a_1, S_2=a_1+a_2, \dots) \text{ So } S_n=\sum_{k=1}^{n} a_k \text{ for n=1,2,...}$$

• If the sequence of partial sums converges, then the infinite series converges.
• If the sequence of partial sums diverges, then the infinite series diverges.

This tells us that we must determine a formula for the sequence of partial sums and its limit in order to evaluate the infinite series. This is where Geometric Series are useful.

A Geometric Series is made up of a sequence of partial geometric sums. Now what exactly is a geometric sum? A geometric sum is a finite sum where each term is a constant multiple of the term before it. The terms in a geometric sum with $n$ terms take the form: $$S_n=a+ar+ar^2+ar^3+\dots+ar^{n-1}=\sum _{k=0} ^{n-1} ar^k$$ Where $a\neq 0$ and r=real numbers.

For example, the geometric sum where n=5, a=0.75, and r=0.5 is: $$0.75+(0.75*0.5)+(0.75*0.5^2)=0.75(0.5+0.25)= \\ \sum _{k=0}^4 ar^k$$

Now we want to find the value of the geometric sum. This can be done using the following formula. $S_n=a\frac{1-r^n}{1-r}$ $\\$ These geometric sums form the sequence of partial sums needed to form a geometric series. The value of the series can be found by taking the limit of the equation above. i.e. $$\lim\limits_{n \to \infty}a\frac{1-r^n}{1-r}$$ This limit can be found by examining the behavior of $r^n$ as $n$ approaches infinity. $\\$ $\bullet \text {When } \lvert r \rvert > 1 \text{ the limit does not exist and the series diverges}$ $\\$ $\bullet \text {When } \lvert r \rvert = 1 \text{ the series becomes } \sum_{k=0}^{\infty}a=a+a+a+\dots \text {,which diverges} \\ \bullet \text {When } \lvert r \rvert < 1 \text{ the geometric converges to } \frac{a}{1-r} \\ \bullet \text{ When } \lvert r \rvert =-1 \text { the series also diverges }$

The infinite geometric series $\sum_{k=0}^{\infty} ar^k = a + ar + ar^2 + ar^3 + \dots$ converges if $|r|<1$, but it diverges if $a\neq0$ and $|r| \geq 1$. If the series converges, then it equals $\frac{a}{1-r}$ precisely.