Question

How can you tell if an infinite geometric series diverges or converges?

Answer 1

I just don't understand how to tell, I know what they are...

Answer 2

The common ratio between successive terms must be between -1 and 1 (excluding -1 and 1).

Think of it this way: If I multiply a given number (say, the initial term of the geometric series - let's use \(a=2\) as an example) by a number between 0 and 1, the result would be smaller.

For example, 50% of 2, i.e. \(\dfrac{1}{2}\times2\), gives you 1. Taking half again, you have \(\dfrac{1}{2}\times1=\dfrac{1}{2}\). Next is \(\dfrac{1}{4},\dfrac{1}{8}\) and so on.

Now suppose the common ratio is greater than 1, say 1.5. This is the same as taking 150% of the first number, i.e. \(\dfrac{3}{2}\times2=3\), which is larger than the first term 2. Multiplying again, you have \(\dfrac{3}{2}\times3=\dfrac{9}{2}=4.5\), which is greater than the previous term of 3.

Answer 3

So for the general form of the geometric series with common ratio \(r\), \(\sum\limits_{k=0}^\infty ar^k\) converges if \(|r|<1\), and diverges if \(|r|>1\).

Answer 4

ohhhh

Answer 5

I have another question...

Answer 6

How can you find the sum of an infinite geometric series?

Answer 7

Provided that you know the series converges, you have
\[\sum_{k=0}^\infty ar^k=\frac{a}{1-r}\]

Answer 8

does k have to equal 0?

Answer 9

because this is what I have: n=1 and then beside the sigma is -4(1/3)^n-1

Answer 10

Not necessarily. The starting index can be any finite number. I happen to prefer this form of the formula because it's easier to remember.

If we wanted the value of the (convergent) series starting at \(k=2\), for instance, here's what we could do:
\[\begin{align*}\sum_{k=2}^\infty ar^k&=\sum_{k=0}^\infty ar^k-a-ar\\\\
&=\frac{a}{1-r}-a-ar\\\\
&=\frac{a-a(1-r)-ar(1-r)}{1-r}\\\\
&=\frac{ar^2}{1-r}
\end{align*}\]
In general, you have
\[\sum_{k=N}^\infty ar^k=\frac{ar^N}{1-r}\]

Answer 11

ok... now that confused me xp sorry... do you think you can explain it more?

Answer 12

There's not much more to explain here... Are you confused by one of the steps above?

Answer 13

yes.

Answer 14

Which one?

Answer 15

the second and third one...

Answer 16

so if k=1, would it just be \[\frac{ a }{ 1-2 }-a\]?

Answer 17

oops, the 2 should be an r

Answer 18

\[\begin{align*}\sum_{k=2}^\infty ar^k&=\color{blue}{\sum_{k=0}^\infty ar^k}\color{red}{-a-ar}\\\\
&=\color{blue}{\frac{a}{1-r}}-a-ar\\\\
&=\frac{a-a(1-r)-ar(1-r)}{1-r}\\\\
&=\frac{ar^2}{1-r}
\end{align*}\]
The red terms in the first line are the first two terms in the series starting at \(k=0\). We want to subtract them because they are already accounted for in the blue series starting at \(k=0\). This subtraction gives a way to write the \(k=2\) series in terms of the \(k=0\) series. The advantage to doing this is that we already have a formula for the blue series, which is the blue expression in the second line.

The next line is some algebraic work. I multiplied the two red terms by \(\dfrac{1-r}{1-r}\) so that they had the same denominator as the blue term. This allows me to combine the numerators as in the third line. The last line is simplification.

Answer 19

And yes, if \(k=1\) then you only subtract that first term \(a\).

Answer 20

ohhhhhhhhhhhhhhh... Okay. It all makes sense now. Thanks for your help. I have one more question, if you don't mind..

Answer 21

This may be a dumb question, but what does it mean if a series converges to 10?

Answer 22

That means all the terms in the series add up to 10.

Here's one such series:
\[\sum_{k=0}^\infty 5\left(\frac{1}{2}\right)^k=\frac{5}{1-\frac{1}{2}}=10\]
Another:
\[\sum_{k=0}^\infty 9\left(\frac{1}{10}\right)^k=\frac{9}{1-\frac{1}{10}}=10\]

Answer 23

oh, okay. Thanks! :)

Answer 24

@SithsAndGiggles I don't think I did it correctly... The answer doesn't make sense..

Answer 25

nevermind. I figured out what I did wrong.