3
$\begingroup$

Does the condition $\lim\limits_{n\to\infty}\frac{a_{n+1}}{a_n}=1$ imply that $a_n$ convergent?

When I say convergent I mean to $L\in \mathbb{R}$ or to $\infty$ or to $-\infty$

I have a gut feeling this statement is false, just can't find a series which contradicts it.

Thanks a lot

$\endgroup$
  • $\begingroup$ @Plopperzz It goes to infinity. $\endgroup$ – Mark Feb 21 at 21:38
  • $\begingroup$ It goes to infinity.. read the first line in my question $\endgroup$ – Shaq Feb 21 at 21:38
  • 3
    $\begingroup$ Here is a counter-example: math.stackexchange.com/a/3015738/42969. $\endgroup$ – Martin R Feb 21 at 21:43
  • $\begingroup$ Ok great Martin this a fine contradiction! I am happy my gut feeling didn't disappoint me $\endgroup$ – Shaq Feb 21 at 21:48
7
$\begingroup$

A counter-example is $a_ n = 2 + \sin(\log(n))$ which is not convergent (it oscillates between $1$ and $3$). But $$ \left\lvert \frac{a_{n+1}}{a_n} - 1 \right\rvert = \left\lvert \frac{\sin(\log(n+1))- \sin(\log(n))}{2 + \sin(\log(n))} \right\rvert \\ \le \left\lvert \sin(\log(n+1))- \sin(\log(n)) \right\rvert = \left\lvert \frac{\cos(\log(x_n))}{x_n}\right\rvert $$ for some $x_n \in (n, n+1)$, using the mean value theorem. It follows that $$ \left\lvert \frac{a_{n+1}}{a_n} - 1 \right\rvert \le \frac{1}{n} \to 0 \, , $$ i.e. $\frac{a_{n+1}}{a_n} \to 1$.

Roughly speaking, $(a_n)$ oscillates, but with decreasing frequency, so that the ratio of successive sequence elements approaches one.

(This example is taken from $\lim\limits_{n \to \infty} \frac{a_n}{a_{n+1}} = 1 \Rightarrow \exists c \in \mathbb{R}: \lim\limits_{n \to \infty} {a_n} = c$. I copied it here because the other question excludes sequences converging to $\pm \infty$, so it is not an exact duplicate.)

$\endgroup$
3
$\begingroup$

A counterexample: $a_n = \exp(\sin\sqrt{n})$. The ratio $\frac{a_{n+1}}{a_n}$ is $$\frac{a_{n+1}}{a_n}=\exp(\sin(\sqrt{n+1})-\sin(\sqrt{n}))\approx \exp\frac{\cos(\sqrt{n})}{2\sqrt{n}}\approx 1+\frac{\cos(\sqrt{n})}{2\sqrt{n}}\to 1$$ but $a_n$ oscillates between $e^1$ and $e^{-1}$, not converging to anything or diverging to $\infty$.

$\endgroup$
1
$\begingroup$

Hint

Think to a sequence $(a_n)$ such that $\lim\limits_{n \to \infty} (a_{n+1}-a_n)=0$, but that oscillates infinitely between $1$ and $2$.

The limit points of $(a_n)$ is the segment $[1,2]$ and $\lim\limits_{n \to \infty} \frac{a_{n+1}}{a_n}=1$.

Some examples here and there.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.