Question

If $\left|r\right|<1$, the sum of the geometric series a

$\sum _{k=1}^{\infty }a{r}^{k-1}$is ________ .

Short answer

If$\left|r\right|<1$, the sum of the geometric series a $\sum _{k=1}^{\infty }a{r}^{k-1}$is $\frac{a_1}{(1-r)}$.

Step-by-step solution

Step 1. Given information.

If $\left|r\right|<1$, the sum of the geometric series a $\sum _{k=1}^{\infty }a{r}^{k-1}$is _________.

• A geometric progression, also known as a geometric sequence, is a sequence of non-zero numbers where each term after the first is found by multiplying the previous one by a fixed, non-zero number called the common ratio.

Step 2. Fill in the blanks.

The sum $s_n$of first $n$terms of a geometric sequence is given by

$$\begin{gathered} s_n=\frac{a_1(1-r^n)}{(1-r)} \\ {s}_n=\frac{a_1}{(1-r)}-\frac{a_1{r}^n)}{(1-r)} \end{gathered}$$

If $\left|r\right|<1$then $\left|r^n\right|$approaches $0$as$n\to \infty$then the term$\frac{a_1{r}^n}{(1-r)}$approaches $0$and $s_n$approaches $\frac{a_1}{(1-r)}$as $n\to \infty$.

Therefore, if$\left|r\right|<1$the sum of the geometric series a$\sum _{k=1}^{\infty }a{r}^{k-1}$is$\frac{a_1}{(1-r)}$.