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Topic: infinite series question
Replies: 3   Last Post: Nov 24, 2002 12:18 PM

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 Randall L. Rathbun Posts: 62 Registered: 12/6/04
Re: infinite series question
Posted: Nov 24, 2002 12:18 PM

unrealistic wrote:

> Randall--
>
> Thank you for your repsonse to my probably pretty basic question on
> the infinite series.
>
> Now if r is basically 0<r<1, and if we want the series to be an
> incresing series, then we have to assume something slightly different,
> correct? (I'm not sure I put this right but the series needs to start
> with the smallest number larger than 0 and the progress toward 1 in
> our series).

Yes, the series needs start with an initial term that is not zero.
and 0< r < 1 (more on this later about r)

You bring up the point that we can have a negative term, let's use the
example of the initial term a = -1 and r = 1/3

Then we have the series
-1 -1/3 -1/9 -1/27 -1/81
and you can see that this is similar to a positive series, only it has a
negative sum. You could factor the -1 from the series and get

-1 * ( 1 + 1/3 + 1/9 + 1/27 + 1/81 + ...)
-1 * ( 3/2 )
-3/2

Can r be less than 0? Let's try it and see.

Something interesting happens when we pick -1 < r < 0, say -2/5.
Then we have an alternating series. Let the first term be 1, and we then
have

1 -2/5 4/25 -8/125 16/625 -32/3125 + ..

Does this have a sum? If you work out this series long enough and add the
terms, you will find that the sum is 5/7. (or use the formula)

> The way I understand it, what you suggested works for between 1 and 2
> (right?) but it we are looking at between 0 and 1, it has to be
> different because of the issue with multiplying by 0.

Not necessarily. Suppose we pick a = 1/3 and r = 3/4

This is the series 1/3 1/4 3/16 9/64 27/256 ...

The sum is 4/3

So nothing is really special about a. We can pick a to be something quite
large, say 1920, and r = 1/8

The series is 1920 240 30 15/4 15/32 15/256 15/2048 ...
and the sum is 15360/7 or 2194 + 2/7

So nothing special about a, except that if we pick a = 0, then nothing
happens. So the values to be avoided are a=0 and r=0.

>
> Is there some special sort of notation I am missing here?
>

Nope, in summary the initial term, a can be anything except 0, but -1 < r <
1 except we notice that at r=0, we don't have a series anymore, just the
first term a.

When r >= 1 the series diverges, an example is a=1 and r=2, we get the sum
of powers of 2

1 + 2 + 4 + 8 + 16 + 32 + 64 + 128 + 256 + 512 + 1024 + ...

and the partial sums are 2^n-1, which can be made as big as you like.

and for r <= -1, I am not sure if a sum even exists. Suppose we let a=1 and
r = -2, then we get the alternating series:

1 -2 4 -8 16 -32 64 -128 256 -512 1024 -2048

This has the partial sums of 1,-1,3,-5,11,-21,43,-85,.... which are wildly
diverging, so not much can be said other than the sums oscillate with
increasing amplitude. That's why I've said that I wonder if an infinite sum
even exists.

I hope this sheds some more light on series for you.

Randall

Date Subject Author
11/23/02 unrealistic
11/24/02 Randall L. Rathbun
11/24/02 unrealistic
11/24/02 Randall L. Rathbun