Doable Leaving Cert Limit?

Michael Collins

I remember helping someone out with Leaving Cert Higher Maths a while back, when I came across this question on an arithmetico-geometric series [2007 Paper 1, Q5 (c)(i) and (ii)]:

Part (i):

[latex] \displaystyle {\rm Find}\; S_n\; {\rm of\; sequence\; with\; general\; term}\; nx^n, \; |x| < 1 [/latex]

This is straightforward enough, the answer being:

[latex] \displaystyle S_n = \frac{x(1-x^n)}{(1-x)^2} - \frac{nx^{n+1}}{1-x} [/latex]

But now Part (ii) asks you to, hence, find the sum to infinity of the series. This is where I came across a bit of an issue:

Taking the limit as [latex] n \to \infty [/latex], we have

[latex] \displaystyle \lim_{n \to \infty} S_n = \lim_{n \to \infty} \frac{x(1-x^n)}{(1-x)^2} - \lim_{n \to \infty} \frac{nx^{n+1}}{1-x} = \frac{x}{(1-x)^2} - \lim_{n \to \infty} \frac{nx^{n+1}}{1-x}[/latex]

Now for the rightmost term we have a problem since, while

[latex] x^n \to 0\; {\rm as}\; n \to \infty [/latex], we obviously have [latex] n \to \infty\; {\rm as}\; n \to \infty [/latex] giving us an indeterminate form

[latex] \displaystyle \lim_{n \to \infty} \frac{nx^{n+1}}{1-x} = \frac{\infty. 0}{1-x} [/latex]

which can be solved by repeated applications of l'Hôpital's rule...But I can't think of any Leaving Cert methods that will do it?

Intuitively we can tell that probably [latex] x^n \to 0\ [/latex] faster than [latex] n \to \infty [/latex] and the term vanishes, but that seems a bit wishy washy, even for Leaving Cert. (Although that appears to be exactly what the Marking Scheme does - it doesn't meantion the n part at all in fact!)

Anyone have any ideas?

Fremen

Not sure about the "hence" here. The way I'd normally approach this is to sum the geometric sequence to infinty in x

[latex] \sum_{n=0}^\infty x^n = \frac{1}{1-x} [/latex]

If you differentiate both sides and then multiply both sides by x, the infinite arithmetico-geometric series pops out.

I suppose if you can show that [latex]S_n[/latex] converges, then it follows that the individual terms in the sum tend to zero. Now,

[latex] \frac{nx^{n+1}}{1-x} = nx^{n}\left( \frac{x}{1-x}\right) [/latex]
is just a constant multiple of the nth term of [latex]S_n[/latex], so it must tend to zero. That way, the problem reduces to showing that [latex]S_n[/latex] converges. Is that covered in leaving cert?

Fremen

Double post: The ratio test says S_n converges, so everything works ok.

MathsManiac

On the core LC course, limits are only treated informally. Some basic limits are to be accepted by inspection/investigation, with the caveat that a proper treatment at 3rd level would prove these things. There are no epsilons and deltas and such.

I think in the context of this question, it's expected that people would have "discovered" previously, without proof, that n*(x^n) goes to 0 as n gets large.

I used to teach at 2nd level, and never gave a rigorous proof of this limit, as the relevant tools aren't on the course.

Anyway, here's an alternative to Fremen's proof. Derivatives of infinite series are not on the LC, but then neither is the version I have here. I'm a bit rusty on this stuff, but I think it's ok:

Taking the case where x is positive, we can show that u(n) is eventually monotonic decreasing. (True once n > x/(1-x).)
It's also clearly bounded below by 0.

Thus, it has a limit L.

It also satisfies the recurrence relation. u(n+1) = x * u(n) + x^(n+1).

Taking limits on both sides yields L=x*L, and since 0<x<1, we have L=0.

Similarly, when x is neagtive, u(n) is m.i. and bounded above and the rest is similar.

The case x=0 is trivial.

Even though that's not on the course, I think the two basic tools involved would be deemed plausible by an LC student:
1. that a sequence whose terms keep decreasing, but which has a barrier below that it can't cross, must have a limit.
2. that, if u(n) tends to a limit, then u(n+1) must tend to the same limit.

Whether or not these two things are easier to accept on faith than an assertion that the derivative of an infinite sum of functions is the sum of their derivatives, I'm not sure. (Well, 2 is easy to accept, but I'm thinking of 1 really.)

Fremen

Nice proof. You need to work a little harder when x is negative since U(n) oscillates between positive and negative, but it should easily be fixable.

MathsManiac

Fremen wrote: »

Nice proof. You need to work a little harder when x is negative since U(n) oscillates between positive and negative, but it should easily be fixable.

Oops! Didn't look carefully enough to spot that!

knockraheen

Are any of you guys got an opinion on project maths .From what I hear most teachers dis agree with the draft syllabus and all agree it should not be introduced in 5th year in Sept 2010.
With regard to your APGP these have an interesting history .Asked for the first time on the 1975 LCH paper 2 question 3 .
You were asked to show the sum to infinity of tr^t= r/(1-r)^2 using the fact that the limit as n goes to infinity of n/p^n=o if p> 1 . APGP's have appeared 3 times 1975,2000,and 2007 they are a bit like hayley's comet in 2000 they just wanted the sum to infinity ."On the core LC course, limits are only treated informally" Syllabus 1994.

Fremen

knockraheen wrote: »

Are any of you guys got an opinion on project maths

From the sample paper I've seen, it looks awful. Students going into technical subjects in university will be hit hard by it.