📂Lemmas

Finding the Sum of a Geometric Sequence

Formula

Given a geometric sequence $a_{n} = a r^{n-1}$ with the first term $a$ and common ratio $r$, $$\sum_{k=1}^{n} a_{k}= {{a (1- r^{n} ) } \over {1-r}}$$

Proof

Let’s denote it as $\displaystyle S= \sum_{k=1}^{n} a_{k}$. Then, $$S= a + ar + \cdots + ar^{n-2} + ar^{n-1}$$ Multiplying both sides by $r$ gives $$rS= ar + a r^2 + \cdots + ar^{n-1} + ar^{n}$$ Subtracting the two equations from each other, $$S - rS = (1-r)S = a- a r^n$$ Dividing both sides of the resulting equation by $1-r$, $$S=\sum_{k=1}^{n} a_{k}= {{a (1- r^{n} ) } \over {1-r}}$$

■

Explanation

Unlike the sum of an arithmetic sequence, this formula itself is very commonly used. The proof method is slightly different but it’s not so complicated that it requires additional study.

Geometric series are also called Geometric Series. When people often say something has “increased geometrically,” that’s what they’re referring to. Thus, most people who are not familiar with mathematics are using the term “geometrically” incorrectly.

What happens if the term $n$ in a geometric series grows infinitely? If $|r|<1$, it will converge, and if $|r|>1$, it will diverge. This consideration of $n \to \infty$ in a geometric series is referred to as an “infinite geometric series.”

Infinite Geometric Series

When $|r|<1$, $$\sum_{n=1}^{\infty} a r^{n-1} = { a \over {1-r}}$$

Since $n \to \infty$ leads to $ar^n \to 0$, it naturally derives from the geometric series.