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Let $⟨a_n⟩$ be a sequence of real numbers such that $a_n \neq 0$. Suppose $\lim_{n\rightarrow\infty}a_{n+1}/a_n = 1$. Can this sequence be oscillatory? If not, how do we prove this?

NB: By oscillating, I mean sequences that do not diverge properly (i.e they do not go to positive or negative infinity). For instance, a sequence like $⟨1,2,1,2,1,2,\dots⟩$ or $n·(-1)^n$.

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They can oscillate, in the way you describe. Take, for example, the sequence: $$a_n = 2 + \cos(\pi \cdot \log_2 n).$$ Note that $a_{2^m} = 2 + (-1)^m$, so the sequence cannot be converging. But, since $|a_n| \le 3$, nor can it diverge to $\pm \infty$. I believe this matches your definition of "oscillating".

Now, we need to show $a_{n+1}/a_n \to 1$ as $n \to \infty$. To show this, we extend $a_n$ to the corresponding continuous, differentiable function on $(0,\infty)$: $$f(x) = 2 + \cos(\pi \cdot \log_2 x).$$ It has derivative: $$f'(x) = -\frac{\pi}{x \ln 2} \sin(\pi \cdot \log_2 x)$$ Mean value theorem on $[n, n+1]$ shows that there is some $x_n \in (n, n+1)$ such that $$-\frac{\pi}{x_n \ln 2} \sin(\pi \cdot \log_2 x_n) = f'(x_n) = f(n+1) - f(n) = a_{n+1} - a_n.$$ Dividing both sides by $a_n$, $$\frac{a_{n+1}}{a_n} - 1 = -\frac{\pi \sin(\pi \cdot \log_2 x_n)}{x_n \ln 2 \cdot (2 + \cos(\pi \cdot \log_2 x_n))}.$$ Given that $\sin$ and $\cos$ are contained between $-1$ and $1$, and that $n < x_n < n+1$, we can therefore conclude that: $$\left|\frac{a_{n+1}}{a_n} - 1\right| = \left|\frac{\pi \sin(\pi \cdot \log_2 x_n)}{x_n \ln 2 \cdot (2 + \cos(\pi \cdot \log_2 x_n))}\right| \le \frac{\pi }{x_n \ln 2} \le \frac{\pi }{n \ln 2}.$$ By squeeze theorem, $\left|\frac{a_{n+1}}{a_n} - 1\right| \to 0$, i.e. $\frac{a_{n+1}}{a_n} \to 1$.

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There is a more direct sort of answer. Often you can transform a "multiplicative" requirement into an "additive" one using the exponential function. In this case, if $b_n$ is an oscillatory sequence such that $b_{n + 1} - b_n \to 0$, then $a_n = e^{b_n}$ will be an example.

Sequences like this $b_n$ are closely related to the fact that $c_n \to 0$ is not a sufficient condition for $\sum_n c_n$ to converge. If you pick any positive sequence $c_n$ tending to zero, but such that $\sum_n c_n$ tends to $\infty$, then you can multiply the terms by $\pm 1$ to make the series oscillate between $0$ and $1$, say. Then you can take $b_n$ to be the partial sums.

Explicitly, some examples of such $c_n$ are $1/n$, or $\sqrt{n + 1} - \sqrt n$, or $\log(n + 1) - \log n$. A very nice example is \begin{equation*} (c_n)_n = (1, \tfrac 12, \tfrac 12, \tfrac 13, \tfrac 13, \tfrac 13, \tfrac 14, \tfrac 14, \tfrac 14, \tfrac 14, \tfrac 15, \dotsc) \end{equation*} so the sequence is made up of terms $1/k$, where $1/k$ appears $k$ times. Then we get \begin{equation*} (b_n)_n = (0, 1, \tfrac 12, 0, \tfrac 13, \tfrac 23, 1, \tfrac 34, \tfrac 24, \tfrac 14, 0, \tfrac 15, \tfrac 25, \dotsc) \end{equation*} and \begin{equation*} (a_n)_n = ( e^{0}, e^{1}, e^{\tfrac 12}, e^{0}, e^{\tfrac 13}, e^{\tfrac 23}, e^{1}, e^{\tfrac 34}, e^{\tfrac 24}, e^{\tfrac 14}, e^{0}, e^{\tfrac 15}, e^{\tfrac 25}, \dotsc). \end{equation*}

But now that we are thinking about sequences where the difference between adjacent terms tends to zero, we might notice that often if two numbers have a small absolute difference, then their ratio will be approximately $1$. In particular, if there is some $M > 0$ such that $|b_n| \ge M$ for all $n$, then the sequence $b_n$ itself will actually give an example, because \begin{equation*} |b_{n + 1}/b_n - 1| = |(b_{n + 1} - b_n)/b_n| \le M^{-1}|b_{n + 1} - b_n| \to 0. \end{equation*}

So another example of such a sequence is given by adding $1$ to the sequence $b_n$ above: \begin{equation*} (a_n)_n = (1, 2, \tfrac 32, 1, \tfrac 43, \tfrac 53, 2, \tfrac 74, \tfrac 64, \tfrac 54, 2, \tfrac 65, \tfrac 75, \dotsc). \end{equation*} Essentially the shapes we're looking for is a "zig-zag" that gets less steep as time goes on. The slick $a_n = 2 + \cos(\pi \log_2 n)$ answer is also of this form - you just have to do a bit more thinking to make sure you've got all the properties you want. For this one you can also deduce that $a_{n + 1} / a_n \to 1$ by looking at the differences, by the way!

Indeed, clearly $\log_2(n + 1) - \log_2 n \to 0$. Then because $\cos$ is uniformly continuous (eg because it's Lipschitz), we have $a_{n + 1} - a_n \to 0$. Also, clearly $a_n \ge 1$ for all $n$, so we are done.

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Yes this is possible, depending on what you mean by oscillatory

If you maintain that the values "oscillate" around the value 0, then this isn't possible, since we would then have

$$ sgn(T_{n+1}) = -sgn(T_{n})$$

For any term in the sequence, and so the limit of the sequence can be at best $-1$

But if you consider sequences oscillating around any any value, then something like

$$ T_n = \frac{(-1)^n}{n^2} + 1 $$ works, since it converges on 1

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  • $\begingroup$ By oscillating, I mean sequences that do not diverge properly (i.e they do not go to positive or negative infinity). For instance, a sequence like <1,2,1,2,1,2..> or $n*(-1)^n$ $\endgroup$ Jun 3, 2022 at 6:49
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    $\begingroup$ Your sequence converges to $1$. A convegent sequence is never considered oscilaltory. $\endgroup$ Jun 3, 2022 at 7:22
  • $\begingroup$ Thank you, I didn't know this $\endgroup$
    – Carlyle
    Jun 3, 2022 at 12:46
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Here is another example. Let

$$ a_n = e^{f(n)}, \qquad \text{where } f(x) = x^{1/4}\sin(\tfrac{\pi}{2} x^{1/2}). $$

Then by the mean value theorem, there exists $\xi \in (n, n+1)$ such that

\begin{align*} \left| f(n+1) - f(n) \right| = \left| f'(\xi) \right| &= \left| \frac{1}{4\xi^{3/4}}\sin(\tfrac{\pi}{2}\xi^{1/2}) + \frac{\pi}{4\xi^{1/4}}\cos(\tfrac{\pi}{2}\xi^{1/2}) \right| \\ &\leq \frac{1}{4n^{3/4}} + \frac{\pi}{4n^{1/4}}, \end{align*}

and hence $f(n+1) - f(n) \to 0$ by the squeezing theorem. So,

$$ \lim_{n\to\infty} \frac{a_{n+1}}{a_n} = \lim_{n\to\infty}e^{f(n+1) - f(n)} = 1. $$

On the other hand, it is clear that $\liminf_{n\to\infty} a_n = 0$ and $\limsup_{n\to\infty} a_n = \infty$.

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