Question

consider the infinite geometric series.in this image the lower limit of the summation is "n=1"a. Write the first four terms of the series.b. Does the series diverge or converge? c. If the series has a sum, find the sum.

Answer 1

The geometric series is;

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

when n=1, we have

$s_1=-4(\frac{1}{3})^{1-1}=-4(\frac{1}{3})^0=-4\times1=-4$

When n=2, we have,

$s_2=-4(\frac{1}{3})^{2-1}=-4(\frac{1}{3})^1=-4\times\frac{1}{3}=-\frac{4}{3}$

When n = 3, we get,

$s_3=-4(\frac{1}{3})^{3-1}=-4(\frac{1}{3})^2=-4\times\frac{1}{9}=-\frac{4}{9}$

when n = 4, we get,

$s_4=-4(\frac{1}{3})^{4-1}=-4\times(\frac{1}{3})^3=-\frac{4}{27}$

a. So, the first four terms of the series are: - 4, -4/3, -4/9 and -4/27

The sum to infinity of the series is:

$S_{\infty}=\frac{a}{1-r}=\frac{-4}{1-\frac{1}{3}}=\frac{-4}{\frac{2}{3}}=\frac{-4\times3}{2}=-2\times3=-6$

b. The series, as we can see, is CONVERGENT, because it has a definite value as its sum.

c. The sum of the series is - 6