A look at a few Tripos questions III

Here’s another one.

10F. State without proof the Integral Comparison Test for the convergence of a series \sum_{n=1}^\infty a_n of non-negative terms.

Determine for which positive real numbers \alpha the series \sum_{n=1}^\infty n^{-\alpha} converges.

In each of the following cases determine whether the series is convergent or divergent:

(i) \displaystyle \sum_{n=3}^\infty \frac 1{n\log n},

(ii) \displaystyle \sum_{n=3}^\infty \frac 1{n\log n(\log\log n)^2},

(iii) \displaystyle \sum_{n=3}^\infty \frac 1{n^{1+1/n}\log n}.

I don’t know exactly what was referred to in the course as the integral comparison test, but since all the sequences being summed are monotone decreasing I’ll go for a neat statement that assumes that (and coincides with what Wikipedia refers to as the integral test).

Let f be a decreasing real-valued function defined on the interval [1,\infty). Then the series \sum_{n=1}^\infty f(n) converges if and only if the integral \int_1^\infty f(t)dt is finite.

Now the second part, which is pretty standard, and works very quickly if you use the integral test. I’m not sure there’s much more to say, so here’s the answer.

For every \alpha>0 the function f(x)=x^{-\alpha} is decreasing. The value of \int_1^x t^{-\alpha}dt is \log x when \alpha=1, and [t^{1-\alpha}/(1-\alpha)]_1^x otherwise. When \alpha<1, this equals (x^{1-\alpha}-1)/(1-\alpha), which tends to infinity. [We could have argued instead by comparing with the \alpha=1 case.] When \alpha>1, it equals (1-x^{1-\alpha})/(\alpha-1), which tends to 1/(\alpha-1). Therefore, the series converges if and only if \alpha>1.

We now have our first dilemma. The series we’re being asked to investigate go from 3 to \infty instead of from 1 to \infty. How much should we say about this? Should we give a more complicated statement of the integral test that allows series that start at n=3?

Since this is a minor matter — after all, the convergence of a series is not affected by the first few terms — I suggest treating it in a minor way. That doesn’t mean ignoring it completely, but it means just indicating in a minimal way to the examiner that you are aware of the difficulty and know how to deal with it. If you do that, then the examiner cannot reasonably remove any marks. You might, for example, write this. “Since the convergence of a series is not affected by its first few terms, it is clearly enough in the examples below to find a function defined on the interval [3,\infty) instead.”

Note that I haven’t even said exactly what I mean (which would have taken longer), but I’ve written something that I couldn’t have written if I didn’t know what was going on. I stress that that kind of short cut is appropriate only if what you are discussing is a side issue and not the main point of the question.

Right, let’s try to deal with these series. I myself prefer grouping terms to using the integral test — I feel that it tells me why the series converges or diverges rather than zapping it with a magic formula — but the question clearly intends me to use the integral test (even if it doesn’t actually insist on it), so I’d better do that unless it gets really horrible.

Hmm, can I integrate 1/x\log x? Yes, because it is of the form g(h(x))h'(x): I take g(x)=1/x and h(x)=\log x. I just need something that differentiates to g, and \log x will do very nicely. So the function that differentiates to 1/x\log x is \log\log x.

I’ll confess that I know (from the grouping terms approach) that the series \sum_n 1/n\log n tends to infinity like \log\log n, and that that helped me get slightly more quickly to the observation in the previous paragraph. Anyhow, time to write something.

(i) Let f(x)=1/x\log x. Then

\int_3^x f(t)dt=[\log\log t]_3^x=\log\log x-\log\log 3.

Since this tends to infinity, the series diverges.

What about the next part? Does anything differentiate to the function 1/x\log x(\log\log x)^2? It looks horrible.

But this is the Cambridge Tripos, and it’s a pure maths question. So it probably isn’t going to involve a horrible calculation. This is a very important clue: it means that it is a good idea to see whether we can find something that works easily.

A rather natural idea is to use the chain rule, since that worked last time. Aha! We’ve just seen that \log\log x differentiates to 1/x\log x, so we can use the chain rule here. We just need something that differentiates to 1/u^2. Well, -1/u does the job, so probably -1/(\log\log x) is going to do the job. Quick check: yes, that does indeed differentiate to what we want.

(ii) Let f(x)=1/x\log x(\log\log x)^2. Then \int_3^xf(t)dt=[-1/\log\log t]_3^x=1/\log\log 3-1/\log\log x. This converges to 1/\log\log 3. Therefore, the series converges.

Now for the third part. Oh dear, that’s a disgusting looking function. How on earth are we supposed to integrate that?

Here again, don’t forget that you are doing a pure maths Tripos question. It is incredibly unlikely that your examiner will expect you to do a horrible integral. (This is not just for your sake, but also your examiner’s — marking a large pile of questions that involve a lot of calculation is only marginally preferable to having your fingernails extracted one by one.) So let us use that information. It probably means that the intended method of solving this part is not integrating a function that almost certainly doesn’t have an integral in closed form.

Incidentally, some people might have been tempted to apply that argument to the function 1/x\log x(\log\log x)^2, but I think that in the context — we were clearly going to apply the integral test at least once, managed to do so for the first part, and observed there that the derivative of \log\log x is 1/x\log x — it looked a lot less nasty than it would do if you were just presented it out of the blue.

Going back to this part, if you’ve decided that actually integrating 1/x^{1+1/x}\log x isn’t possible, what else are your options for solving this problem? Well, another great law of mathematics is use what you already know. This law applies with particular force in Tripos questions, where “what you already know” often translates to “what you have done earlier in the question”. Is there anything from earlier in the question that we might be able to use?

Well, all three functions we’re supposed to look at are small modifications of 1/n\log n. In the first part, the “small” modification was in fact the zero modification, and the series diverged. In the second part, we made it very slightly smaller, by dividing by (\log\log n)^2, and it converged. So let’s try to think of how the third function relates to 1/n\log n. And that is easy to see: it divides it by n^{1/n}. So the question we’re facing is whether dividing by n^{1/n} does enough to tip the series over from being one that diverges to being one that converges.

Obviously if we want to decide about that, we would like some idea of what n^{1/n} does as n\to\infty.

A useful tip for many such questions: if you can’t immediately see the answer, take logs. You need a bit of common sense about when to apply this tip, but here, since we’re raising something to a power, taking logs has a good chance of simplifying matters.

The log of n^{1/n} is \log n/n. What does that do? It tends to 0. (That’s a good example of something that’s more basic than what you’re being asked to prove, and therefore legitimate to assume.) Therefore, n^{1/n} tends to 1. But in that case it’s very unlikely that dividing by n^{1/n} makes any difference.

How can we prove quickly that it makes no difference? With convergence and divergence there are two principles that can be used to produce quick proofs — in conjunction with the comparison test. They are

(1) if you change finitely many terms in a series, it makes no difference to whether that series converges


(2) if you multiply the terms of a series by a positive constant, it makes no difference to whether that series converges.

Armed with those two principles, we have the following very short argument (which is obviously a special case of a more general fact).

(iii) Since \log n/n\to 0 as n\to\infty, n^{1/n}\to 1. Therefore, n^{1/n}\leq 2 for all sufficiently large n. It follows that 1/n^{1+1/n}\log n\geq 1/2n\log n for all sufficiently large n. Therefore, by part (i) and the comparison test, \sum_{n=1}^\infty 1/n^{1+1/n}\log n diverges.

Note that I could have argued that the function x^{1/x} is bounded, and even tried to calculate a bound. But that would have involved thought and time and gained me an extra zero marks. Better to write something crude and get on to a new question.

A final remark is that if you pick out the bits of what I wrote above that were supposed to form part of the final answer, you’ll end up with something that can be written out in about three minutes (though obviously you’ll also have to spend some time thinking of the answers). That is often the case with Tripos questions: they are designed to be easy if you’re on top of the material, and quite difficult otherwise. As with the last question, I think this one was quite easy as Tripos questions go, but I can see that the n^{1+1/n} factor might have led a number of people astray.

I’d like to end this post by explaining how I’d tackle a convergence question of another kind that sometimes comes up. What can we say about the convergence of the series

\displaystyle \sum_{n=1}^\infty \frac{n^2+8n+100}{(n^2-35/2)^2}?

I’m not concerned here about the answer, but about how to write a very quick and fully rigorous answer. The answer, incidentally, is obviously that it converges, since the terms decrease roughly like 1/n^2. However, the nth term is bigger than 1/n^2, so it’s oversimplifying a bit too much to say, “By comparison with 1/n^2, the series converges,” even if that is more or less the right reason.

Here’s my cheating method. You just write down some very easy inequalities that hold when n is sufficiently large. For example, if n\geq 10, then n^2\geq 8n and n^2\geq 100, so n^2+8n+100\leq 3n^2. Also, if n\geq 10, then n^2-35/2 is certainly at least n^2/2, so (n^2-35/2)^2\geq n^4/4.

Here’s what I’d actually write for the question. It’s quick and ugly and that’s what I like about it.

If n\geq 10, then n^2+8n+100\leq 3n^2 and (n^2-35/2)^2\geq n^4/4. It follows that the nth term is at most 12/n^2. Therefore, by the comparison test and the fact that the first nine terms do not affect convergence, the series converges.

If your version of the comparison test is a for-n-sufficiently-large version, then you don’t even have to say, “and the fact that the first nine terms do not affect convergence.” I’m not sure what your lecturer went for.

It’s tempting to waste time worrying about things like whether you could get just as good an inequality for smaller n, or whether you could improve the constant 12 when n\geq 10. But there’s no need. You just care about whether the series converges.

If you were very short of time, you could take a gamble and just guess that the number one million is going to work, which, unless the numbers in the question are huge, it will. A safe option would be this.

If n\geq 10^6, then the nth term is at most 10^6/n^2. Therefore, by the comparison test, the series converges.

A slightly riskier option is this.

For each n, the nth term is at most 10^6/n^2. Therefore, by the comparison test, the series converges.

In general, I don’t recommend this sort of joke approach if you’ve got a few seconds to spare, because if you make the inequality true by miles, you haven’t demonstrated to the examiner all that convincingly that you can back up the statement you are making. With the earlier version, where I said that if n\geq 10 then n^2+8n+100\leq 3n^2, the examiner can see where the number 10 came from (it was to make n^2 equal to 100), so it’s clear that I haven’t just cheated and written something down that was bound to be true and not checked it.

5 Responses to “A look at a few Tripos questions III”

  1. Max Says:

    There is a typo in your statement of the comparison test: your summation symbol doesn’t have a summand (it should be f(n)).

    Thanks very much — corrected now.

  2. Gareth Says:

    My “cheating” method is the following. We pull out the leading term on the top and bottom, then use that the remaining terms tend to a constant, and so we can then bound them above or below as appropriate. (If we’re trying to bound above, then we bound the leftover factor on top above, and that on the bottom below. If we’re trying to bound below, then reverse that.)

    E.g., we have n^2+8n+100 = n^2(1+8/n+100/n^2), and (n^2-35/2)^2 = n^4(1-35/2n^2)^2. Since 1+8/n+100/n^2 \to 1, then is some N such that it’s less than 2 for all n>N. Similarly, since (1-35/2n^2)^2 \to 1, there’s some N' such that it’s greater than 1/2 for all n>N'.

    Therefore, for all n>\max(N,N'), we have the summand less than 4/n^2.

    (If we’d had n^3+8n+100 on top, say, then we’d point out that 1+8/n^2+100/n^3 is eventually greater than 1/2, and that (1-35/2n^2)^2 is eventually less than 2, making the ratio eventually greater than 1/4n.)

    I like this because it doesn’t require thinking of any bounds at all, and it fits well with the comments about “the terms decrease roughly like 1/n^2“.

    • gowers Says:

      I like that approach too — in fact, it’s pretty similar to the way I decided to handle the series in part (iii).

    • Greg Martin Says:

      Why do it in two steps when you can do it in one? The quotient $(1+8/n+100/n^2)/(1-35/2n^2)^2$ tends to $1$, hence is less than $2$ for sufficiently large $n$.

    • Gareth Says:

      Indeed. I was more giving my version of the “cheating method” in the post, in which top and bottom were attacked separately. It can certainly be made simpler, as you suggest.

Leave a Reply

Fill in your details below or click an icon to log in:

WordPress.com Logo

You are commenting using your WordPress.com account. Log Out /  Change )

Twitter picture

You are commenting using your Twitter account. Log Out /  Change )

Facebook photo

You are commenting using your Facebook account. Log Out /  Change )

Connecting to %s

%d bloggers like this: