TITLE: Divergent sequence where $\lim_{n\to\infty}(a_{n+p}-a_n) = 0 $
QUESTION [6 upvotes]: I need to show that $(a_n)=\sin(\ln n)$ is a bounded sequence that has no limit, but for which $$\lim_{n\to\infty}(a_{n+p}-a_n )= 0\; \forall p \in \mathbb N.$$
I can show that the sequence is bounded and that it diverges, but I'm stuck with the other part. So far I've found that
$\lim_{n\to\infty} (\ln (n+p)-\ln n) = 0$, but I don't really know where to go next. The easy way to get the sines in would of course be
$$ \lim\ln(n+p)-\lim\ln n = 0
\Leftrightarrow \lim\sin(\ln(n+p))-\lim\sin(\ln n)=0 \Leftrightarrow \lim(\sin (\ln (n+p)) - \sin (\ln n)) = 0,$$
but as both $\sin x$ and $\ln x$ diverge, I'm not sure that'd be ok.
So am I anywhere near the right direction or should I try a different approach?

REPLY [2 votes]: Let $\varepsilon>0$ and $p\geq 1$ be fixed. Since $\sin(x)$ is a uniformly continuous function, given $\varepsilon>0$ there is a $\delta>0$ such that if $|x-y|<\delta$ then $|\sin(x)-\sin(y)|<\varepsilon$. Since $\ln(n+p) - \ln n = \ln (\frac{n+p}{n}) \to 0$ as $n\to \infty$, there is a number $N=N(p)$ such that if $n\geq N$ then $|\ln(n+p)-\ln n|<\delta$. Now take $x=\ln(n+p)$ and $y=\ln n$. We conclude that:

Let $p\geq 1$ be fixed. For any $\varepsilon>0$ there is a number $N=N(p,\varepsilon)$ such that $$|a_{n+p}-a_n|=|\sin(\ln(n+p))-\sin(\ln n)|<\varepsilon.$$

Thus, by the definition of limit, we have shown that $\lim_{n\to \infty} (a_{n+p}-a_n) = 0$, as desired.