Question

Consider the infinite geometric series ..

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.

Walk Me Through Please !

Answer 1

\[\sum_{n = 1}^{\infty} -4(\frac{ 1 }{ 3 })^(n-1)\]

Answer 2

It looks like you meant to say\[\sum_{n=1}^\infty -4\left(\dfrac{1}{3}\right)^{n-1}\]
The way your series is written suggests it should like this, that is.

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

So Part A. - 4 , -1.33 , -.44 , -.14 ? @SithsAndGiggles

Answer 4

Yes, though I'd round up -0.14 to -0.15, since you -4/27 = -0.148148... .

Answer 5

@SithsAndGiggles so it shouldnt have a sum since its an infinite geometric series

Answer 6

That's only the case when the ratio you're given does not satisfy |r|<1. Does that look familiar?

Answer 7

So There Is A sum ? But Idk How To FInd That

Answer 8

Okay, I'll go back a bit. You should have this somewhere in your class notes, if you have those around.
For a given geometric series of the form
\[\sum_{n=1}^{\infty}ar^{n-1},\]
the series converges if |r|<1. Think about it: if a = 1 and r = 1/2, for example, you have a series that looks like this:
1(1/2)^(1-1) + 1(1/2)^(2-1) + 1(1/2)^(3-1) + ∙∙∙
1 + 1/2 + 1/4 + ∙∙∙
Adding up these first few numbers, you get 1.75.
Adding more terms of the series, you get closer and closer to 2. It may not be apparent, but the series indeed converges to 2.

Now consider a = 1 and r = 2, as another example. The first few terms, added, are
1(2)^(1-1) + 1(2)^(2-1) + 1(2)^(3-1) + ∙∙∙
1 + 2 + 4 + ∙∙∙
Each successive term here is greater than the preceding one, so this series diverges.

The general conclusion to draw from this small examination is that a given geometric series converges if |r|<1 (i.e., -1 < r < 1), and diverges if |r|≤-1 or |r|≥1.

If it converges, it has a sum. There's a formula for geometric series with |r|<1 that you can easily use. Do you know of it?

Answer 9

No sorry

Answer 10

\[\sum_{n=1}^{\infty}ar^{n-1} = \dfrac{a}{1-r}\]