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As far as I am aware, $\displaystyle \int_1^\infty \displaystyle \frac{1}{\sqrt{x}} \mathrm d x$ diverges due to the $p$ test, meaning that the series $\displaystyle \sum_{x=1}^\infty \displaystyle \frac{1}{\sqrt{x}}$ also diverges (integral comparison test). But this would mean that $\displaystyle \lim_{x\to\infty}{\frac{1}{\sqrt{x}}}=0,$ which isn't true for a diverging series?

EDIT: Isn't there a theorem about how the series $\displaystyle \sum_{x=0}^\infty a_n$ converges for a sequence $a_n$ if $\displaystyle \lim_{n\to\infty} a_n=0$?

EDIT #2: I've found the caveat to this theorem, so I suppose the explanation about the rate of increase of $\displaystyle \frac{1}{\sqrt{x}}$ suffices

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  • $\begingroup$ Divergent things can have terms that go to 0, they just don't go to 0 fast enough. $\endgroup$ – Randall Nov 9 '17 at 4:02
  • $\begingroup$ Sure it is true... there are a lot of series that diverge to 0 but their sum goes to infinity; think of the harmonic series (sum of 1/n for all n) $\endgroup$ – E-A Nov 9 '17 at 4:03
  • $\begingroup$ @Randall I'm aware, but this seems to conflict with the theorem that I mentioned in my edit. $\endgroup$ – user98937 Nov 9 '17 at 4:19
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    $\begingroup$ There is no such theorem. $\endgroup$ – Randall Nov 9 '17 at 4:22
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    $\begingroup$ If $\sum_n a_n$ converges then $\lim_n a_n=0$. The converse is false. All dogs are mammals but horses exist. $\endgroup$ – Randall Nov 9 '17 at 4:23
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A sum like $\displaystyle\sum_1^\infty \frac{1}{n^p}$ is called a $p$-series (for "power" law), and it converges for $p>1$, but diverges for $p<1.$ For the boundary case $p=1,$ the series is called the harmonic series, which also diverges. Powers higher than one make fractions below one get smaller, which enhances convergence, while powers less than one get those fractions bigger, closer to one, which does not help convergence. So it's not too surprising that there is some cutoff below which the power laws do not converge, despite going to zero termwise.

Although your question was about the integral, not the sum, it turns out that the convergence of the $p$-series sum $\displaystyle\sum_1^\infty \frac{1}{n^p}$ and the integral $\displaystyle\int_1^\infty \frac{dx}{x^p}$ is the same. I hope you don't mind the shift in context.

So the upshot is, it's not enough that the function go to zero. A reciprocal power law only converges if its power is greater than one, so that it's going to zero faster than $\frac{1}{n}$.

So how do we reconcile this fact with the statement:

the series $\displaystyle \sum_{x=0}^\infty a_n$ converges for a sequence $a_n$ if $\displaystyle \lim_{n\to\infty} a_n=0$?

Well the statement, as written, is not correct. For example, the harmonic series has terms $a_n=\frac{1}{n}$ with $\lim a_n = 0,$ and yet the sum $\sum^\infty\frac{1}{n}$ and integral $\int^\infty\frac{dx}{x}$ do not converge. So where did you get the idea? Well, the inverse (if a statement is an implication $p\to q$, then its inverse is $\neg p\to\neg q$) of this statement is a theorem, sometimes called the divergence test

If $\displaystyle \lim_{n\to\infty} a_n\neq 0,$ then $\displaystyle \sum_{x=0}^\infty a_n$ does not converge.

Going to zero termwise is not sufficient to guarantee convergence, but failing to go to zero is sufficient to guarantee divergence.

And in general, the truth of a statement $p\to q$ does not guarantee the truth of the converse $q\to p$ nor the inverse $\neg p \to \neg q.$ Although it is a common mistake to assume that they do. One statement which you can conclude is the contrapositive: $\neg q\to\neg p.$ Thus an alternate true statement we can make is:

If $\displaystyle \sum_{x=0}^\infty a_n$ converges, then it follows that $\displaystyle \lim_{n\to\infty} a_n = 0.$

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  • $\begingroup$ Yeah...but it seems to conflict with the theorem that I mentioned in my edit. $\endgroup$ – user98937 Nov 9 '17 at 4:22
  • $\begingroup$ @user98937 let me address in an edit of my own $\endgroup$ – ziggurism Nov 9 '17 at 4:22
  • $\begingroup$ thanks, I managed to revisit the theorem's definition again and saw the flaw in the reasoning, but your answer was equally helpful $\endgroup$ – user98937 Nov 9 '17 at 4:31

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