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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.

Consider The Infinite Geometric Seriesin This Image The Lower Limit Of The Summation Is N1a Write The First Four Terms Of The Seriesb Does The Series Diverge Or class=

Sagot :

The geometric series is;

[tex]s_n=\sum ^{\infty}_{n\mathop=1}-4(\frac{1}{3})^{n-1}[/tex]

when n=1, we have

[tex]s_1=-4(\frac{1}{3})^{1-1}=-4(\frac{1}{3})^0=-4\times1=-4[/tex]

When n=2, we have,

[tex]s_2=-4(\frac{1}{3})^{2-1}=-4(\frac{1}{3})^1=-4\times\frac{1}{3}=-\frac{4}{3}[/tex]

When n = 3, we get,

[tex]s_3=-4(\frac{1}{3})^{3-1}=-4(\frac{1}{3})^2=-4\times\frac{1}{9}=-\frac{4}{9}[/tex]

when n = 4, we get,

[tex]s_4=-4(\frac{1}{3})^{4-1}=-4\times(\frac{1}{3})^3=-\frac{4}{27}[/tex]

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:

[tex]S_{\infty}=\frac{a}{1-r}=\frac{-4}{1-\frac{1}{3}}=\frac{-4}{\frac{2}{3}}=\frac{-4\times3}{2}=-2\times3=-6[/tex]

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

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