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Let $(u_n)_n$ be an unbounded (i.e. $(|u_n|)_n$ is not bounded from above) sequence of real numbers such that $ \frac{u_{n+1}}{u_n} \to_{n\to\infty} 1 $. Does this necessarily imply that $(u_n)_n$ diverges to $+\infty$? I can't find any counter-examples, so I think this might be true. Using Cesaro's mean sequence, we get that $\sqrt[n]{u_n} \to_{n\to\infty}1$, which leads to the fact that $n=o(\ln(u_n))$. So, if we write $u_n=f(n)$, $f$ would be a polynomial or weaker than a polynomial (in the sense of limits in infinity) like $\ln$ and $\ln \circ \ln$, and more generally $\ln \circ \dots \circ \ln$. I feel like my counter-example function should be a very "weak" function that is unbounded but still doesn't tend to infinity, it seems to me intuitively that this function lies in the limit between bounded functions and weak unbounded functions (such as $\ln$), so the problem would be to show that there is no such limit function, or that there is. (Although, I think my reasoning is a little bit flawed as sequences are discrete and by translating the problem to functions, I ignore this whole property.)

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    $\begingroup$ What about $u_n = -n$ with $u_n \to -\infty$? $\endgroup$ – Martin R Dec 13 '20 at 21:55
  • $\begingroup$ What if we impose the further condition that $u_n$ has to be positive? $\endgroup$ – Peanut Dec 13 '20 at 22:04
  • $\begingroup$ @MartinR by unbounded, I meant from above and from below. Else, we can impose that $u_n$ is positive just like Peanut said, since we need when taking the $n$-root. $\endgroup$ – Ansper Dec 13 '20 at 22:13
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A simple counterexample would be $u_n = -n$.

But it is wrong even if we require that all $u_n$ are positive. A counterexample is $$ u_n = 1 + n^{1/4} \sin^2(n^{1/2}) $$ which is unbounded, but also approaches the value $1$ infinitely often.

With $f(x) = x^{1/4} \sin^2(x^{1/2})$ we have $$ \left| \frac{u_{n+1}}{u_n} - 1\right | \le |f(n+1) - f(n)| = |f'(c_n)| $$ for some $c_n \in (n, n+1)$, and that converges to zero because $$ f'(x) = \frac 14 x^{-3/4} \sin^2(x^{1/2}) + x^{-1/4} \sin(x^{1/2}) \cos (x^{1/2}) \to 0 $$ for $x \to \infty$.

Roughly speaking, the idea is to construct a sequence which oscillates with increasing amplitude, but decreasing frequency.

Another way (inspired by this) is to start with the sequence $$ \begin{align} x_n =\; & 0, 1, \\ & 0, \frac 12, \frac 22, \frac 32, 2, \frac 32, \frac 22, \frac 12, \\ & 0, \frac 13, \frac 23, \ldots, \frac 83, 3, \frac 83, \ldots, \frac 13, \\ & 0, \frac 14, \ldots \, . \end{align} $$ For $k=1, 2, 3, \ldots$ we have “segments” where $x_n$ increases from $0$ to $k$ in increments of $1/k$, and back to $0$ in decrements of $-1/k$. $(x_n)$ is unbounded with $\lim_{n\to \infty} (x_{n+1} - x_n) = 0$. Every non-negative real number is a limit point of this sequence.

Then define $u_n = e^{x_n}$.

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