kpreid | Spot the bogus proof [updated] (Reply)

(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.

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Spot the bogus proof [updated] (Reply)

Spot the bogus proof [updated]

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