Question

Geometric series. In this image, the lower limit of the summation notation is “n=1”.

Answer 1

Solution:

Given the infinite geometric series below

$\sum_{n\mathop{=}1}^{\infty}-4(\frac{1}{3})^{n-1}$

a) For the first four terms, n = 1, 2, 3, 4

Where, n = 1,

$=-4(\frac{1}{3})^{1-1}=-4(\frac{1}{3})^0=-4(1)=-4$

Where, n = 2

$=-4(\frac{1}{3})^{2-1}=-4(\frac{1}{3})^1=-4(\frac{1}{3})=-\frac{4}{3}$

Where, n = 3

$=-4(\frac{1}{3})^{3-1}=-4(\frac{1}{3})^2=-4(\frac{1}{9})=-\frac{4}{9}$

Where, n = 4

$=-4(\frac{1}{3})^{4-1}=-4(\frac{1}{3})^3=-4(\frac{1}{27})=-\frac{4}{27}$

Hence, the first four terms are

$-4,-\frac{4}{3},-\frac{4}{9},-\frac{4}{27}$

b) To confirm if the series converges or diverges,

$\begin{gathered} r=\frac{second\text{ term}}{first\text{ term}}=\frac{-\frac{4}{3}}{-4}=-\frac{4}{3}\times-\frac{1}{4}=\frac{1}{3} \\ r=\frac{1}{3} \end{gathered}$

Since, the common ratio, r is between -1 and 1,

Hence, the series converges

c) To find the sum of the series, we will apply the sum to infinity formula, which is

$\begin{gathered} S_{\infty}=\frac{a}{1-r} \\ Where \\ a\text{ is the first term} \\ r\text{ is the common ratio} \end{gathered}$$\begin{gathered} S_{\infty}=\frac{-4}{1-\frac{1}{3}}=\frac{-4}{\frac{2}{3}}=-4\div\frac{2}{3}=-4\times\frac{3}{2}=-6 \\ S_{\infty}=-6 \end{gathered}$

Hence, the sum is -6