Proving a sequence converges to 0 Let $a_n$ be a sequence of real numbers such that $$\lim_{n \to +\infty}n\Big(\Big|\frac{a_n}{a_{n+1}}\Big|-1\Big)=l \in (0,+\infty)$$ Prove that $a_n$ converges to $0$. I could find that $|\frac{a_n}{a_{n+1}}|$ approaches $1$ as $n$ approaches $+\infty$. That can be deduced from the fact that $$\forall\epsilon>0,\exists n_\epsilon \in \mathbb{N}, \forall n>n_\epsilon, l-\epsilon<n\Big(\Big|\frac{a_n}{a_{n+1}}\Big|-1\Big)<l+\epsilon$$
Thus $$\frac{l-\epsilon}{n}+1<\Big|\frac{a_n}{a_{n+1}}\Big|<\frac{l+\epsilon}{n}+1$$
If we take the limit as $n$ approaches $+\infty$ we have $$\lim_{n \to +\infty}\Big|\frac{a_n}{a_{n+1}}\Big|=1$$
I know I haven't used the fact that $l>0$, I only used the fact that $l$ is finite. How should i go about proving $a_n$ approaches $0$?
 A: First, because $l>0$, then there exists $N \in \mathbb{N}$ such that for every $n \geq N$, one has
$$n \left( \left| \frac{a_n}{a_{n+1}}\right|-1\right) > 0$$
Hence for every $n \geq N$, one has $ \left| \dfrac{a_n}{a_{n+1}}\right|-1 > 0$, so $|a_n| > |a_{n+1}|$. So the sequence $(|a_n|)_{n \geq N}$ is decreasing, and since it is positive, it converges to a limit $l' \geq 0$.
Let's suppose that $l' > 0$. One has $$n \left( \left| \frac{a_n}{a_{n+1}}\right|-1\right) \sim l, \quad \quad \text{so} \quad \left| \frac{a_n}{a_{n+1}}\right|-1 \sim \frac{l}{n}, \quad \quad \text{so} \quad |a_n|-|a_{n+1}| \sim \frac{l\times l'}{n}$$
Because the series $\displaystyle{\sum \frac{l\times l'}{n}}$ diverges, you get than by comparison, the series $\displaystyle{\sum (|a_n|-|a_{n+1}|)}$ also diverges, which means that the sequence $(|a_n|)$ diverges. Contradiction.
So $l'=0$ and you are done.
A: Since
$$\lim_{n \to +\infty}n\Big(\Big|\frac{a_n}{a_{n+1}}\Big|-1\Big)=l \in (0,+\infty)$$
for fixed $\epsilon=\frac l2$, $\exists N\in \mathbb{N}$ such that, for $n\ge N$
$$ \frac l2=l-\epsilon<n\Big(\Big|\frac{a_n}{a_{n+1}}\Big|-1\Big)<l+\epsilon =\frac32l$$
or
$$ \frac n{n+\frac{3l}{2}}<\Big|\frac{a_{n+1}}{a_n}\Big|<\frac{n}{n+\frac l2}. $$
From this, one has, for $n>N$,
$$ |a_{n}|\le|a_N|\Pi_{k=N}^n\frac{k}{k+\frac l2}. $$
Noting the series
$$ \sum_{k=1}^\infty\ln\frac{k}{k+\frac l2}=-\sum_{k=1}^\infty\ln\frac{k+\frac l2}k=-\sum_{k=1}^\infty\ln(1+\frac{l}{2k})=-\infty $$
so one has
$$ \lim_{n\to\infty}\Pi_{k=N}^n\frac{k}{k+\frac l2}=0$$
and hence
$$ \lim_{n\to\infty}a_n=0. $$
