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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.

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    $\begingroup$ Have you tried Gauss's test? $\endgroup$
    – Simply Beautiful Art
    Commented Aug 13, 2017 at 13:45
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    $\begingroup$ @SimplyBeautifulArt This would simply be taking the result for granted, no? $\endgroup$
    – Did
    Commented Aug 13, 2017 at 13:48
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    $\begingroup$ 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? $\endgroup$
    – Did
    Commented Aug 13, 2017 at 13:50
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    $\begingroup$ 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? $\endgroup$
    – Did
    Commented Aug 13, 2017 at 14:11
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    $\begingroup$ @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. $\endgroup$
    – Did
    Commented Aug 13, 2017 at 14:33

2 Answers 2

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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.

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    $\begingroup$ Would the downvoter care to explain his/her downvote? $\endgroup$
    – Jack D'Aurizio
    Commented Aug 13, 2017 at 14:51
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    $\begingroup$ Very unique answer. +1 the evaluation of infinite product in closed form is another gem. $\endgroup$
    – Paramanand Singh ♦
    Commented Aug 13, 2017 at 14:57
  • $\begingroup$ @JackD'Aurizio Do they ever? $\endgroup$
    – Clement C.
    Commented Aug 13, 2017 at 15:02
  • $\begingroup$ @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. $\endgroup$
    – Jack D'Aurizio
    Commented Aug 13, 2017 at 15:09
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    $\begingroup$ @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). $\endgroup$
    – Clement C.
    Commented Aug 13, 2017 at 15:23
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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.

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