Question

HELP PLEASE! Have been having tons of trouble with infinite geometric series, hopefully someone can help! Attachment below.

Answer 1

First step: find the common ratio.

Answer 2

I believe it is 3.

Answer 3

@whpalmer4

Answer 4

Okay, let me ask you a question: if the common ratio is 3, that means that we get each new term by multiplying the old one by 3. How then do some of our terms end up being positive numbers, and others end up being negative numbers?

Answer 5

Blahhhh I don't know ): I asked someone else on here once and they said it was 3, so I guess I automatically assumed they were right :/ Bad mistake. How would I find the common ratio?

Answer 6

Well, the "common ratio" should suggest that there's a ratio in common between the terms, no? What is the second term divided by the first term? How about the third term divided by the second term?

Answer 7

3 would be the right answer if all of the terms were positive, or all of them were negative.

Here's a hint:

suppose we start with 1
-1*1 = -1
-1*-1 = 1
-1*1 = -1
-1*-1 = 1
...

Answer 8

Wait, so all I have to do is divide the second term by the first?

Answer 9

yes!

Answer 10

Would the first term be 3? Maybe that's why I thought it was the common ratio.

Answer 11

if you aren't certain that you have a geometric sequence, then you'd want to try the same test with the 3rd and 2nd terms to see if you get the same answer (if you don't, it isn't geometric)

Answer 12

terms here are 3, -1, 1/3, -1/9, 1/27, -1/81 (more terms than they gave you, but that's what they'll be)

Answer 13

what do you have to multiply 3 by to get -1?
what do you have to multiply -1 by to get 1/3?
what do you have to multiply 1/3 by to get -1/9?

actually I was incorrect in saying that 3 would be the common ratio if they were all positive or all negative, but there is a 3 involved!

Answer 14

-3?

Answer 15

mmm, no. but close :-)

3*-3 =

Answer 16

Ohhhh. -9!

Answer 17

3*-3 = -9. We need a multiplier that causes our numbers to get smaller, not larger. In particular, for the thing to converge to a sum, we need the common ratio to have a magnitude less than 1:

\[|r|<1\]or\[-1<r<1\]

That way each each term adds a smaller and smaller piece.

Have you ever heard of Zeno's paradox? There are a number of forms of it, but it boils down to summing an infinite geometric series. To walk across the room, you must first walk halfway across the room. But to walk halfway across the room, you must first walk half of halfway, or one quarter of the way across the room. But to do that ...

Yet we have no trouble walking across the room, unless it's a room in my house, which is so full of stuff that you can't see the floor to walk across it :-)

Answer 18

We need a different number than -3 as our common ratio, because our terms are shrinking in size, not growing. How about a fraction?

\[3*r = -1\]\[-1*r = \frac{1}{3}\]Solving either one of those for \(r\) should give us the correct common ratio, \(r\)

Remember, each term in the series is obtained by multiplying the previous term by the common ratio. 3 is our first term, -1 is our second term, so \(3*r= -1\).

Answer 19

I see we are at the time of day where OpenStudy starts being flaky, displaying responses out of order, long pauses, etc. :-( Let's hope we finish up before it completely loses it!

Answer 20

\[3*r = -1\]\[\frac{3*r}{3} = \frac{-1}{3}\]\[r = -\frac{1}{3}\]

As a check, we'll multiply \(-1\) by \(r\) to see if we get the next term:

\[-1*(-\frac{1}{3}) = \frac{1}{3}\checkmark\]Good, we've found the common ratio, and it is \[r =-\frac{1}{3}\]Now to find the sum, we plug it into our formula:
\[S = \frac{a}{1-r}\]where \(a = 3\) (the first term in the series) and \(r\) has the value we just found.

What is the result?

Answer 21

As an illustration, I computed the first 12 terms and the running sum (first as a fraction, then as a decimal):

\[\begin{array}{cccc}
a_n & S_n & a_n & S_n \\ \hline \\
3 & 3 & 3. & 3. \\
-1 & 2 & -1. & 2. \\
\frac{1}{3} & \frac{7}{3} & 0.333333 & 2.33333 \\
-\frac{1}{9} & \frac{20}{9} & -0.111111 & 2.22222 \\
\frac{1}{27} & \frac{61}{27} & 0.037037 & 2.25926 \\
-\frac{1}{81} & \frac{182}{81} & -0.0123457 & 2.24691 \\
\frac{1}{243} & \frac{547}{243} & 0.00411523 & 2.25103 \\
-\frac{1}{729} & \frac{1640}{729} & -0.00137174 & 2.24966 \\
\frac{1}{2187} & \frac{4921}{2187} & 0.000457247 & 2.25011 \\
-\frac{1}{6561} & \frac{14762}{6561} & -0.000152416 & 2.24996 \\
\frac{1}{19683} & \frac{44287}{19683} & 0.0000508053 & 2.25001 \\
-\frac{1}{59049} & \frac{132860}{59049} & -0.0000169351 & 2.25 \\
\end{array}\]

You can see that each contribution to the sum is rather smaller than the one before (necessarily so, or the answer would be infinity!) and that the sum homes in on the final answer, bouncing back and forth between being slightly too small and slightly too large, until finally the difference is smaller than the precision of the computer's arithmetic.

Answer 22

I got the answer, 9/4!