34.3k views
0 votes
Need help ASAP!!!
Find the sum of each infinite geometric series, if it exists​

Need help ASAP!!! Find the sum of each infinite geometric series, if it exists​-example-1
User Karla
by
7.8k points

1 Answer

0 votes

The first series diverges because each successive term is getting multiplied by -3, so the common ratio is larger than 1 in magnitude.

Assuming the second series is

3 - 3/2 + 3/4 - 3/8 + ...

then it converges because the common ratio between terms is -1/2, which is less than 1 in magnitude.

To find the value of the sum, let S denote the n-th partial sum, i.e. the sum of the first n terms in the series:

S = 3 - 3/2 + 3/4 - 3/8 + ... + 3(-1/2)^(n - 2) + 3(-1/2)^(n - 1)

Multiply both sides by -1/2:

(-1/2) S = -3/2 + 3/4 - 3/8 + 3/16 - ... + 3(-1/2)^(n - 1) + 3(-1/2)^n

When we subtract S and (-1/2) S, we see all the middle terms canceling, and we can solve for S:

S - (-1/2) S = 3 - 3(-1/2)^n

S/2 = 3[1 - (-1/2)^n]

S = 6[1 - (-1/2)^n]

As n gets larger, the (-1/2)^n term will converge to 0, leaving us with

lim (n -> infinity) S = 6

so that the series converges to 6.

User Chadrik
by
8.9k points

No related questions found

Welcome to QAmmunity.org, where you can ask questions and receive answers from other members of our community.