Question

What is the formula for the sum of an infinite geometric series?

Answer 1

A sequence of non-zero numbers is called a geometric progression, if the ratio of a term and the term preceding it is always a constant quantity. That ratio is called the common ratio of the G.P. We will derive the formula for the sum of n terms of a geometric progression and then in that formula we will apply the limit of n tending to infinity.

Complete answer:
Let ${S_n}$ denote the sum of n terms of a G.P with first term $a$ and common ratio $r$. Then we can write as;
${S_n} = a + ar + a{r^2} + ...... + a{r^{n - 1}}.......(1)$
Now multiplying both sides by $r$, we get;
$r{S_n} = ar + a{r^2} + ...... + a{r^{n - 1}} + a{r^n}......(2)$
Subtracting $(2)$ from $(1)$ we get;
${S_n} - r{S_n} = a - a{r^n}$
Because the rest all terms are common in both and will cancel each other.
Taking common we get;
$\Rightarrow {S_n}\left( {1 - r} \right) = a\left( {1 - {r^n}} \right)$
On shifting we get;
$\Rightarrow {S_n} = \dfrac{{a\left( {1 - {r^n}} \right)}}{{\left( {1 - r} \right)}}$
Here, $r$ should not be equal to $1$.
Now this is the sum of n terms of a GP.
$\Rightarrow {S_n} = \dfrac{a}{{1 - r}} - \dfrac{{a{r^n}}}{{1 - r}}$
Now suppose $|r| < 1$ i.e., $- 1 < r < 1$.
Now as $- 1 < r < 1$, so, if we increase the power of $r$, then its value will decrease. When we put the power $n \to \infty$, then ${r^n} \to 0$.
Let the sum of infinite terms of this GP be $S$. So, we get;
$\Rightarrow S = \mathop {\lim }\limits_{n \to \infty } {S_n}$
Putting the value, we get;
$\Rightarrow S = \mathop {\lim }\limits_{n \to \infty } \left( {\dfrac{a}{{1 - r}} - \dfrac{{a{r^n}}}{{1 - r}}} \right)$
Now we know, $n \to \infty$, then ${r^n} \to 0$ because $- 1 < r < 1$.
So, we get;
$\Rightarrow S = \mathop {\lim }\limits_{n \to \infty } \left( {\dfrac{a}{{1 - r}} - \dfrac{{a \times 0}}{{1 - r}}} \right)$
On simplification we get;
$\Rightarrow S = \dfrac{a}{{1 - r}}$, $- 1 < r < 1$.

Note:
A convergent sequence is the one which has a finite and unique value. The geometric progression will be convergent only when $|r| < 1$. This means we will get a finite and unique value for the infinite series only when $|r| < 1$. That is why we have applied for this case. If this is not the case then, the series will be divergent and we will not get a finite value.