Not enough information? at Lessons taught; Lessons learnt

Not enough information?

It’s been a busy term, and it’s not over yet, but I promised that I would post something this week. So, a question for you to explore:

The second term of a geometric sequence is 2. What’s the sum to infinity?

If your first thought is ‘it could be anything’, then you’re not alone, but you’re also not correct!

If you get an answer algebraically, can you give a geometric interpretation?

14 Responses to 'Not enough information?'

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I’m sure I’m missing something, but I can only get it down to 2/(r(1-r)), where r < 1 and r != 0 or 1. I get different outcomes if I try different values for r.

Kate Nowak

7 Jun 09 at 12:52 pm
Well, it would seem that the sum would be >= 8, (and 8 only when r=1/2 and g(0) = 4)

I’m ignoring negative terms since you said something about geometry, but I don’t see it?

Pat Ballew

7 Jun 09 at 2:40 pm
I’m with @Pat: ignoring negative terms, the sum can be anything >= 8 (with negative starting values, the sum can be anything less than -1). I think one can show that the only integer values the sum can take are 8 (initial term 4) or 9 (initial term 3 or 6). Of course, that doesn’t seem to be the original question. Also, I don’t seem to have anything to say about the geometric interpretation.

Nick Hamblet

7 Jun 09 at 5:29 pm
Thank you for the comments, which I find interesting for several reasons.

First, I agree with your answer — that you can’t get anything between -1 and 8. Probably the easiest way to show this is to try and solve $s=\frac{2}{r(1-r)}$ for $r$ , and look at the discriminant.

Just stop for a minute and ask yourself if it seems surprising that you can’t get any sum. It certainly surprised me when I first started looking at geometric sequences! If we’re told that the *first* term is 2, then there’s no such restriction on the sums. This immediately makes me want to look at what happens when I fix the third term, or the fourth …

Dragging myself back to earth, I suppose we could stop now we’ve ’solved’ the problem, but that seems very unsatisfactory. I don’t feel like I could justify properly from the algebra *why* 8 (rather than, say, 6) turns out to be the minimum possible positive sum, or why we can’t get -0.9 but we can get -1.1.

If we graph the sum against the ratio, some other features inherent in the algebra become apparent. A ratio of (for example) 0.1 gives us the same sum as a ratio of 0.9. We can see *that* it is true from the algebra, but *why* is it true?

This leads into my second response to your answers, which is a bit meta-ry. I asked this: “If you get an answer algebraically, can you give a geometric interpretation?” Perhaps you weren’t expecting it, but this was a genuine question: I don’t have a geometric interpretation for any part of this — I want one, because it feels like there should be one. After I’ve marked the set of books in front of me tonight, I’ll be looking for one! The meta-y bit: What was it about the setting which made you think that this wasn’t a genuine question? Also, does knowing that *I* don’t have an answer make you more or less interested in looking for one yourself?

(Sorry if this sounds a little wishy-washy, but it’s getting toward the end of the school year, which always makes me a little reflective!)

Jon Ingram

7 Jun 09 at 7:21 pm
Wow, I interpreted your question to be “What’s the one possible value of the sum,” not “What are the restrictions on the value of the sum.” I agree that a geometric interpretation would be gratifying.

Kate Nowak

8 Jun 09 at 10:19 am
Kate, apologies — I see how my poor phrasing would give that impression. It would be better phrased as “What can we say about the sum to infinity?”, I suppose, but that seems a little formal!

Jon Ingram

8 Jun 09 at 10:37 am
Probably I’m just catching up to you folks, but…

For a positive real number s, consider all of the geometric series with second term s, and define f(s) to be the starting term of the series with smallest value. So, from above, f(2)=4. More generally, I think one finds that f(s)=2s. This means that the ratio of this ’smallest’ series is 1/2.

I think we can interpret this as follows: For a geometric series to converge, we know it has a common ratio between terms that is less than 1 in absolute value. So if our second term is s, and we are ignoring series with negative terms, then the first term is any number bigger than s. If we pick a number just slightly bigger than s, the series will converge very slowly, and to a big number. As the value of the first term moves away from s, getting bigger and bigger, the common ratio in the series gets smaller and smaller, and so the dominating terms in the series are the first few (I guess that’s always true, in a sense…). The best balance between a series that converges fast enough to not have a lot of bulk “in the tail”, but yet doesn’t start with too big of an initial term (the value of the series will always be bigger than first_term+s), is apparently a series with starting term 2s, and common ratio 1/2.

In my mind, I wouldn’t have initially guessed 1/2 would be the best. It doesn’t seem “natural” enough. Not that it’s unreasonable: it does split the interval (0,1), in which the ratio lies, pretty nicely. I tried to justify 1/2 to myself another way though, and came up with the following (which is essentially hogwash). Since the ratio is less than 1, we can write it as 1/(1+r’), for some positive real value, r’. Now there is a “natural” real number in (0,inf), namely 1, which gives a ratio of 1/2.

Fascinating question. Thanks Jon.

Nick Hamblet

9 Jun 09 at 11:45 am
Not sure if this constitutes geometry, but one of those things that pops up in stats a lot is that if 0<r<1, then r(1-r) has a max at r=1/2 and the value is 1/4… This shows up all the time when you are trying to estimate the smallest sample size needed to build a confidence interval of a given width… since the standard deviation of a binomial event is sqrt(np(1-p) we just assume the worst case (biggext std dev) and assume p(1-p) = 1/4 and proceed from there…
So here is the geometry… the largest rectangular area you can make with a perimeter of two (a semi perimeter of one) is 1/4…

Pat Ballew

9 Jun 09 at 3:51 pm
OK, so it struck me that if we let 2 = the nth term, not the second, then the minimum value of the sum will be when r= (n-1)/n and the minimum value itself will be 2 (n^n)/(n-1)^(n-1) now that’s cute..
(mini analysis) the value of the sum will be (2/(r^n (r-1) )
and will be smallest when the denominator is largest, so we just need to find the maximum of r^n – r^(n-1) which happens when the derivative is zero…or when n r^(n-1) =(n-1) r^(n-2)   so one (uninteresting) solution is at zero, and the other is at r= (n-1)/(n)

so for term number n, the minimum sum would be

n                  sum
2                   2(2^2)/(1^1) = 8
3                  2 (3^3)/(2^2) = 13.5
4                2(4^4)/(3^3) = 512/27
5                2^(5^5)/(4^4) = 6250/256

and other big numbers down the line…
now back to grading finals….

Pat Ballew

9 Jun 09 at 4:07 pm
Pat and Nick,

Thank you for the analysis, and for the geometric connections which I will have to ponder later on.

Now, being evil and dragging you away from your finals, here are some more directions you can take this:

1) Can we find an approximation which will stop us from having to evaluate fractions like $\frac{10^{10}}{9^9}$ ? [Yes we can, and it involves a famous mathematical constant!]

2) What about if we switch the question around? For example: The sum to infinity of a geometric progression is 1. What are the possible values for the 2nd term? [And, which ratio maximizes this value?]

Twisting the investigation around slightly,

3) The sum to infinity of a geometric progression is 1. What ratio maximizes the product of the first three terms?

Looks like I need to sit down and write a proper post on all these questions… and then try to figure out how to integrate them into my A-level teaching next year!

Jon Ingram

10 Jun 09 at 8:22 am
1) Well it would seem like any (pun intended) number would be an upper limit that gets closer as n-> infinity.. (is that the one you mean?)

2) I think I answered that question in an earlier note… the first on June ninth…

3) and I think that is the same as 2) since the product will be the cube of the second term if my mind is doing this right….

I have completely ignored the negatives because that would require paper and pencil and if I’m going to write things down, I may as well record these grades sitting in front of me…

you are building a nice exploration of geometric series…
thanks

Pat Ballew

10 Jun 09 at 3:57 pm
I have created a Geogebra Applet that I believe illustrates the positive case pretty well. (0<r<1)

Let me know what you think http://scottfarrar.googlepages.com/a2is2_.html

(applet wont work after mid June 2009 since Google is removing the capability for custom java. I will have a redirect on the main site)

Scott

10 Jun 09 at 9:18 pm
For 1) Let’s use the magic of Wolfram Alpha which tells us that a good approximation to $\frac{n^n}{(n-1)^{n-1}}$ for large $n$ is $e (n - \frac{1}{2} )$ .

So if the nth term is 2, the sum to infinity is roughly $e (2n - 1)$ . It’s a surprisingly good approximation, being within 0.05 of the true value for $n=3$ .

For 2) and 3) I was (in an oddly cryptic way) thinking about other questions which would lead to the ratio of 1/2 being the best. Incidentally, if you want to maximise the product of the first two terms, then the best ratio is 1/3.

Jon Ingram

10 Jun 09 at 9:33 pm
I realized I mis-spoke about attaining integer values in my first comment. The only integer values you’ll get as a sum, when the initial term is an integer, is 8 or 9. But that’s not particularly interesting. Just thought I’d point out the correction.

Nick Hamblet

11 Jun 09 at 5:10 pm