Question

Infinite Geometric Series

Answer 1

Consider the infinite geometric series \[\sum_{n=1}^{\infty}-4(\frac{1}{3})^{n-1}\] In this image, the lower limit of the summation notation is "n=1".

1) Write the first four terms of the series.
2) Does the series diverge of converge?
3) If the series has a sum, find the sum.

Answer 2

here.
Part (a) is simple enough. Just plug in n=1,2,3,4 into the sequence and evaluate. For n=1, for example, you have -4(1/3)^(1-1) = -4(1) = -4.
Part (b) requires examining the ratio of the series, 1/3. What do you know about the convergence conditions of geometric series?
Part (c): if the series converges (which you may or may not determine in part (b)), is there a certain formula you learned about the sum of a geometric series?

Answer 3

I got part one.....I got
\(\large -4,~-\frac{4}{3},~-\frac{4}{9},~-\frac{4}{27}\)
correct?

Answer 4

So Part A. - 4 , -1.33 , -.44 , -.14

Answer 5

yep....I get confused on this next part....

Answer 6

what are you confused in

Answer 7

is it converge because the common ratio is less than 1?

Answer 8

then \[\frac{ -4 }{ 1-(\frac{1}{3}) }\]

Answer 9

\[\frac{-4}{\frac{2}{3}}\]
which is -6