If $\frac {a_{n+1}}{a_n} \ge 1 -\frac {1}{n} -\frac {1}{n^2}$ then $\sum\limits_na_n$ diverges

Question

Let $(a_n)_{n \ge 1}$ be a sequence of positive real numbers such that, for every $n\ge1$, $$\frac {a_{n+1}}{a_n} \ge 1 -\frac {1}{n} -\frac {1}{n^2} \tag 2$$ Prove that $x_n=a_1 + a_2 + .. + a_n$ diverges.

It is clear that $x_n$ is increasing, so it has to have a limit. I tried to prove the limit is $+\infty$ but without success. No divergence criteria from series seems to work here.

UPDATE

Attempt: Suppose a stronger inequality holds, namely that, for every $n\ge1$, $$\frac{a_{n+1}}{a_n}\geqslant1-\frac1n \tag 1$$ Then: $$\frac {a_3}{a_2} \ge \frac 1 2\qquad \frac {a_4}{a_3} \ge \frac 2 3\qquad \ldots\qquad \frac {a_{n-1}}{a_{n-2}} \ge \frac {n-3}{n-2}\qquad \frac {a_n}{a_{n-1}} \ge \frac {n-2}{n-1}$$ Multiplying all the above yields $$\frac {a_n}{a_2} \ge \frac 1 {n-1}$$ The last inequality proves the divergence.

@SimplyBeautifulArt This would simply be taking the result for granted, no? — Did, Aug 13 '17 at 13:48
@SimplyBeautifulArt Gauss's test solves the problem for sure, but I would like something rather elementary. — , Aug 13 '17 at 13:48
@Did :-) I suppose so, but if its got a name, you can google the proof. For example, I found this one rather easily. — Simply Beautiful Art, Aug 13 '17 at 13:48
@EugenCovaci that is a contradiction to your statement "No divergence criteria work here": Gauss works. End of Proof. — Marcus Müller, Aug 13 '17 at 13:49
The idea of the proof of Gauss's test is not so complicated, and rather illuminating, I would say. To get to it, how would you prove that $$\frac{a_{n+1}}{a_n}\geqslant1-\frac1n$$ implies divergence? — Did, Aug 13 '17 at 13:50
@Did Writing the inequality for $n=1, 2 ... n$ then multiplying all the resulting inequalities might work. — , Aug 13 '17 at 13:53
Please be more precise (as you guessed, we will adapt the method to your case so we have to know exactly the method in this simpler case). — Did, Aug 13 '17 at 13:55
@MarcusMüller That's nitpicking. The OP obviously meant "none of the most standard (and well-known, widely taught) divergence criteria work here." Otherwise, you can always have one ad hoc divergence criterion per series, and invoke it. — Clement C., Aug 13 '17 at 13:56
Hm, Gauss was one of the convergence criteria for sums I was told in engineering first semester, so I was presuming that it's rather well-known. It's also really easy to remember and motivate, imho, — Marcus Müller, Aug 13 '17 at 14:00
@Did I updated my answer. The problem is (1) is stronger than what I have. — , Aug 13 '17 at 14:02
Excellent. So, you consider $$b_n=\frac1{n-1}$$ then you note that $$\frac{b_{n+1}}{b_n}\leqslant1-\frac1n$$ hence $$\frac{a_{n+1}}{a_n}\geqslant\frac{b_{n+1}}{b_n}$$ hence $$a_n\geqslant Cb_n$$ for some irrelevant but positive value of $C$ and since $\sum b_n$ diverges, you are done. To adapt this strategy to your case, you merely have to find $(c_n)$ such that $\sum c_n$ diverges and $$\frac{c_{n+1}}{c_n}\leqslant1-\frac1n-\frac1{n^2}$$ Any idea? — Did, Aug 13 '17 at 14:11
@Did From (2) doesn't follow $\frac{a_{n+1}}{a_n}\geqslant\frac{b_{n+1}}{b_n}$ — , Aug 13 '17 at 14:15
?? Of course, which is why the sequence $(b_n)$ solving the simpler case $1-\frac1n$ should be replaced by another sequence $(c_n)$ if one wants to solve the case $1-\frac1n-\frac1{n^2}$. — Did, Aug 13 '17 at 14:17
@EugenCovaci Not sure what you mean, but here Did refers to the "simpler" case where you assume $\frac{a_{n+1}}{a_n} \geq 1-\frac{1}{n}$. The idea is to abstract what works in that case, and generalize it to the (tougher, but similar) situation you have. — Clement C., Aug 13 '17 at 14:18
Try something of the form $c_n = \frac{1}{n}+\frac{A}{n^2}$, for some parameter $A$ (this may not be the most elegant). — Clement C., Aug 13 '17 at 14:30
" I've got the idea" Perfect (IMHO, this is by far the most important point). Re $c_n$, did you try some simple form, say $$c_n=\frac1{n+\alpha}$$ for some well chosen value of $\alpha$? When yu will have found $(c_n)$, I suggest that you collect the proof and post it as an answer below. — Did, Aug 13 '17 at 14:30
@Clement C. I beg to differ. That the Gauss test is not "well known and widely taught", is a failure of mathematical education, in the West. In standard text books in Russia (not lost in translation in surprisingly many countries in the East), Gauss test is taught, and rightly so. It would save us from all those annoying "ratio test is inconclusive" questions at MSE, btw. — , Aug 13 '17 at 14:32
@ProfessorVector Right, but the OP was not exposed to this literature, obviously, so we can take this as an opportunity to actually do some maths on this page (something that happens too rarely on MSE), don't you think? You noticed that "doing some maths" neither refers to mentioning that one already knows the result hence there is nothing to prove, nor to "magic" inequalities provided without explanation and yielding the result. — Did, Aug 13 '17 at 14:33
@ProfessorVector Sure, and I agree this ought to be taught (I was taught some variant of it in France, and don't regret it). And so? Unless you have a plan to retroactively change the "mathematical education [system] in the West," this is not really going to help the OP or anyone in the next 10 years or so. Plus, as Did points out... sometimes, getting intuition and proving things from scratch may be a good idea, instead of directly using a sledgehammer. — Clement C., Aug 13 '17 at 14:35
It doesn't satisfy: $\frac{c_{n+1}}{c_n}\leqslant1-\frac1n-\frac1{n^2}$ — , Aug 13 '17 at 14:41
Eugen: You do not "explain" much, sorry to say... Fortunately, @ClementC. has mind-reading abilities. :-) — Did, Aug 13 '17 at 14:43
Ottimo! Now, wrap up everything we said and concoct an answer based on it... :-) — Did, Aug 13 '17 at 14:44
@Did Hey man, I'm a Java programmer trying to help a kid with some math problems. Why are you torturing me? — , Aug 13 '17 at 14:46
Because for some of us, actually doing some mathematics is not a torture. And because a part of the ethos of mathematicians is to be convinced that the whole world, if put in a suitable situation, can do maths and take pleasure from the activity. Yes, optimism... (Will you write an answer?) — Did, Aug 13 '17 at 14:47
Nice post. And, just to be clear: it is perfectly allowed to accept one's own answer. — Did, Aug 13 '17 at 17:16

score 6 · Accepted Answer · answered Aug 13 '17 at 14:25

6

You may notice that $$ \frac{a_{n+1}}{a_n} \geq \left(1-\frac{1}{n}\right)\left(1+\frac{1}{n^2-n-1}\right)^{-1} \tag{1}$$ hence: $$ \frac{a_{N+1}}{a_2}\geq \frac{1}{N}\prod_{n=2}^{N}\left(1+\frac{1}{n^2-n-1}\right)^{-1} \tag{2}$$ but the infinite product $\prod_{n\geq 2}\left(1+\frac{1}{n^2-n-1}\right)^{-1}$ is convergent to a positive number ($-\frac{1}{\pi}\cos\frac{\sqrt{5}}{2}\approx 0.296675134743591$). It follows that $a_{N+1}\geq \frac{C}{N}$ so $\sum_{n\geq 2}a_n$ is divergent.

answered Aug 13 '17 at 14:25

Jack D'Aurizio

353,855

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Would the downvoter care to explain his/her downvote? – Jack D'Aurizio Aug 13 '17 at 14:51
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Very unique answer. +1 the evaluation of infinite product in closed form is another gem. – Paramanand Singh Aug 13 '17 at 14:57
@JackD'Aurizio Do they ever? – Clement C. Aug 13 '17 at 15:02
@ClementC.: it happened once or twice. I have no problem in fixing wrong or not-so-clear parts and I think it does not hurt to request an explanation nicely. – Jack D'Aurizio Aug 13 '17 at 15:09
@JackD'Aurizio I agree, my comment was more of a fatalistic flavor. Downvoters often don't care much about hanging around long enough to explain the rationale, if any. – Clement C. Aug 13 '17 at 15:10
Nice answer, but can you prove $\prod_{n=2}^{\infty}\left(1+\frac{1}{n^2-n-1}\right)^{-1}$ is convergent (or bounded it's enough)? – Aug 13 '17 at 15:17
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@EugenCovaci By comparison, if you don't care about the value. From $$\ln \frac{1}{1+\frac{1}{n^2-n-1}}= \ln\left(1-\frac{1}{n^2}+o\left(\frac{1}{n^2}\right)\right) = -\frac{1}{n^2}+o\left(\frac{1}{n^2}\right)$$ you get $$\ln \prod_{n=2}^N \frac{1}{1+\frac{1}{n^2-n-1}} = \sum_{n=2}^N \ln \frac{1}{1+\frac{1}{n^2-n-1}}$$ which, by comparison, has same convergence nature as $-\sum_{n=2}^N \frac{1}{n^2}$, itself convergent. So the logarithm of the infinite product converges to a finite value: the infinite product itself then does so as well (and the limit is positive). – Clement C. Aug 13 '17 at 15:23
(But knowing Jack D'Aurizio, there is a neat and clever way to get the convergence and the actual limiting value at once.) – Clement C. Aug 13 '17 at 15:26
@EugenCovaci: by exploiting a well-known lemma - If $\prod_{n\geq 2}(1+b_n)$ is an infinite product with $b_n>0$ and $\sum_{n\geq 2}b_n$ converges, then the infinite product is convergent to a positive number. – Jack D'Aurizio Aug 13 '17 at 15:26

score 4 · Answer 2 · edited Aug 13 '17 at 17:21

It's easy to show that, for every $n\ge3$, $$ 1 -\frac {1}{n} -\frac {1}{n^2} \ge \frac {n-2}{n-1}$$ It follows that, for every $n\ge3$, $$\frac{a_{n+1}}{a_n}\ge \frac {n-2}{n-1}$$ Thus, $$\frac {a_4}{a_3} \ge \frac 1 2\qquad \frac {a_5}{a_4} \ge \frac 2 3\qquad \ldots\qquad \frac {a_{n-1}}{a_{n-2}} \ge \frac {n-4}{n-3}\qquad \frac {a_n}{a_{n-1}} \ge \frac {n-3}{n-2}$$

Multiplying all the above, one gets: $$\frac {a_n}{a_3} \ge \frac 1 {n-2}$$ hence $$a_n\ge \frac{a_3}{n-2}$$ The last inequality together with the comparison criteria for series proves the divergence.

If $\frac {a_{n+1}}{a_n} \ge 1 -\frac {1}{n} -\frac {1}{n^2}$ then $\sum\limits_na_n$ diverges

2 Answers2