kpreid | Spot the bogus proof [updated]

(This is (finally, now that the semester is over and I have some free time, heh) one of those posts-related-to-schoolwork I mentioned before.)

Does the infinite series $\displaystyle\sum_{n=2}^{∞} \dfrac{n^2}{n^3 + 1}$ converge?

First, rewrite it to have n = 1 as $\displaystyle -\frac{1}{2} + \sum_{n=1}^{∞} \dfrac{n^2}{n^3 + 1}$ , and discard the constant term since it does not affect convergence. The terms of this series are strictly less than those of $\displaystyle \sum_{n=1}^{∞} \dfrac{n^2}{n^3} = \sum_{n=1}^{∞} \dfrac{1}{n}$ ; therefore, there is some x such that

$\displaystyle \sum_{n=1}^{∞} \dfrac{n^2}{n^3 + 1} < x < \sum_{n=1}^{∞} \dfrac{1}{n}$

Let k be the upper bound of the sum as we take the limit: $\displaystyle \sum_{n=1}^{∞} \dfrac{n^2}{n^3 + 1} = \lim_{k\to ∞}\sum_{n=1}^{k} \dfrac{n^2}{n^3 + 1}$ .

Since $\displaystyle \sum_{n=1}^k \dfrac{1}{n^p}$ is a continuous function of p, for any k there is some p greater than 1 such that

$\displaystyle \sum_{n=1}^k \dfrac{n^2}{n^3 + 1} < x = \sum_{n=1}^k \dfrac{1}{n^p} < \sum_{n=1}^k \dfrac{1}{n}$

Since $\displaystyle \sum_{n=1}^{∞} \dfrac{1}{n^p}$ is a p-series which converges, i.e. has a finite sum, and the series under consideration has a lesser sum by the above inequality, it converges.

Furthermore, the above may be generalized to a proof that any series whose terms are eventually less than those of the harmonic series converges.

However, it is invalid, and in fact $\displaystyle\sum_{n=2}^{∞} \dfrac{n^2}{n^3 + 1}$ diverges.

I managed to convince myself and my calculus teacher with it, but we realized it must be invalid after he presented a counterexample to the general case. I then realized which step was invalid.

You can't use the same k for all three series in the second inequality; each infinite sum has its own independent limit, and what this proof is doing is along the lines of ∞ - ∞ = 0 — assuming that “two infinities are the same size”. Or rather, the inequality itself (among partial sums) is true, but that fact has nothing to do with the properties of the true infinite series.

I would be mildly interested in a more formal description of this sort of failure: how the k inequality is true yet the independent series do not have the same relation.

Update: I have received many informative comments and replied to some; one pointed out an earlier mistake: $\displaystyle \sum_{n=1}^{∞} \dfrac{n^2}{n^3 + 1} < x < \sum_{n=1}^{∞} \dfrac{1}{n}$ is bogus because we can't compare the series until we know they converge.

Flat | Top-Level Comments Only

the issues is that all you know is for each k there is a p such that this inequality is valid. in order to take the limit of this inequality to infinity you would need a p such that this inequality is valid for all k.

When I wrote this, I was thinking of epsilon-delta proofs of limits; for any ε (k) there is a δ (p) such that the inequality holds. Is this not reasonable?

That wouldn't be enough, you'd need some uniformity of p in k.

More specifically, the term you are considering is actually

(http://www.codecogs.com/eqnedit.php?latex=\sum_{n=1}^k \frac 1{n^{p(k)}})

i.e. the p depends on k. Sending k to ∞ could well send this sequence to infinity, unless you have for example a uniform bound like say (http://www.codecogs.com/eqnedit.php?latex=\forall k. p(k) < q < 1) which you don't.

You could also say that the fallacy was to wrongly interchange the limits of k to ∞ and p to 1.

What do you mean by "uniformity of p in k" and "uniform bound", or rather, what term should I look up for more information? I do see that ∀k. p(k) < q < 1 ensures that

is convergent.

Sorry, it should be ∀k. p(k) > q > 1. Such a q is usually called a "uniform" bound because it's independent of k.

But now consider for example

for which there is no such q. We have

for k large enough (because

which means that this kth root will eventually be smaller than 2). Thus, this sum diverges as k goes to infinity, and this is where your proof went wrong.

(Note that the lack of a uniform bound q does not imply that the sum will diverge. There may be some p(k) that approach 1 but slowly enough for the sum to converge.)

Unfortunately, I don't have a good reference for such issues of interchanging limits and uniformity – the Tricki (http://www.tricki.org/) would probably be the right place for this – but they will crop up in many places (interchanging limits, uniform continuity, uniform convergence of continuous functions, interchanging integrals and limits, etc. etc.) and understanding this matter in its many variations is key to understanding calculus; eventually you'll get the hang of it :-).

We have:
sum [n=1..inf] [n^2/(n^3+1)]
forall n>=1, n^2/(n^3+1) < 1/n.

You then claim that:
exists x. sum [1..inf] [n^2/(n^3+1)] < x < sum [1..inf] [1/n]

This step assumes that the left-hand sum is convergent!

When you're comparing the sums of your series and the sum of the harmonic series... you're already assuming both converge -- and in fact they both diverge.

The problem you point to later is much finer than that first problem : finding that one and not the first is very surprising!

PS: you'll probably find wikipedia's page on integral-series comparison (http://en.wikipedia.org/wiki/Integral_test_for_convergence) very helpful.

Ah, I missed the bogus comparison, thanks. I appreciate the importance of not treating undefined/infinite/divergent as if it's finite, but I occasionally fail at actually doing so.

I already know about the integral test and that it shows these series are divergent.

Note that the bogus x is, while bogus, not the main problem of your argument. You can delete all occurrences of x from your proof and still be left with the problem that your term

seems to converge.

n^2/(n^3+1)=1/(n+1/n^2)>=1/(2n) for all n>=1. Use the comparison test.

The idea that you're looking for is uniform/non-uniform continuity. However, one of your fallacies is that that there is a p for each k, and their limit is 1.

I just read a bit about uniform continuity. Is this related to the observation "Looks wrong..." made that my

is actually

exactly...

1/n does not converge, so it does not "limit" the series you are summing... (you only know that it diverges a little bit slower!) mp

Spot the bogus proof [updated]

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Looks wrong right near the start

You're missing the worse problem

Re: You're missing the worse problem

Re: You're missing the worse problem

so many problems...

Re: so many problems...

Re: so many problems...

just to say that