Question

Find the sum of the following infinite geometric series if it exists. (2/5) + (12/25) + (72/125) +...


A. 5/12
B. Does not exist
C. 5/6
D. 6/5

Answer 1

\[\sum_{i=0}^{\infty} (\frac{ 2 }{ 5 })(\frac{ 6 }{ 5 })^{n}\]

This is what your series appears to be. A geometric series in the form of ar^n diverges if r is greater than or equal to 1. Since r is 6/5, meaning greater than one, this series diverges and the sum would be infinite.

Answer 2

So the answer is 6/5 since that is greater than one?

Answer 3

@Psymon

Answer 4

divide the 2nd term by the first term and you get r = (6/5)
thus the sum does not exist at infinity because (6/5)>1

Answer 5

\[\sum_{n=1}^{\infty}(\frac{6}{5})^n\]

Answer 6

no need to start at 0

Answer 7

I've just been taught to start from 0 with geo series. I know it's not necessary, but its habit.

Answer 8

Oh okay, I think I understand what you did there. If I had the following problem, how would I approach it?
Find the sum of the following infinite geometric series if it exists. 1/2 + (-1/4) + 1/8 + (-1/16) +...
It looks similar but I don't know.

Answer 9

if you know its geometric, divide the second term by the first

Answer 10

Alright then. Dividing the 2nd term by the first....

\[\sum_{i=1}^{\infty}(\frac{ -1 }{ 2 })^{n}\]

Since the absolute value of r is between 0 and 1, the sequence converges. If a geo series converges, it converges to the value of a/(1-r), meaning we would have:

\[\frac{ 1 }{ 1-(\frac{ -1 }{ 2 })}\] = (2/3)

Answer 11

\[\frac{\frac{1}{2}}{1-(\frac{-1}{2})}\]

Answer 12

the first term is (1/2)

Answer 13

Yeah, dividing out the first term and ditching it just confused me, lol.

Answer 14

\[\frac{first\space term}{1-ratio}\]

Answer 15

I'm just too used to keeping it there and starting from 0 x_x

Answer 16

This is confusing lol. But the options I have are
A. 1/3
B. 1/5
C. 1/12
D. 1/4

Answer 17

Did we miss a step?

Answer 18

Nah, A is correct, I'm just trying to all of a sudden do something new and messing it up :P zz knows what he's doing.

Answer 19

\[\frac{\frac{1}{2}}{1+\frac{1}{2}}=\frac{1/2}{\frac{3}{2}}=\frac{1}{2}*\frac{2}{3}=\frac{1}{3}\]

Answer 20

just one little issue, but we all make that mistake...they normally start with 1

Answer 21

give psy the medal he did most of it

Answer 22

Oh okay, I forgot about bringing out 3/2 and making it 2/3. Thanks a lot for your help guys!

Answer 23

Yeah, when we learned it once upon a time ago, we always started from 0. Even my textbook starts it from 0.

Answer 24

yeah

Answer 25

Oh well. Good to see something new :P