Divergent sequence where $\lim_{n\to\infty}(a_{n+p}-a_n) = 0$

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?

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This is an interesting example, because it shows that this weak "Cauchy condition" is not enough to guarantee convergence. – Matemáticos Chibchas Mar 19 '13 at 4:46

\begin{align*}\sin \ln (n+p)-\sin \ln n&=2\sin\left(\frac{\ln (n+p)-\ln n}{2}\right)\cos\left(\frac{\ln (n+p)+\ln n}{2}\right)\\[7pt]&=2\sin\left[\frac{1}{2}\ln \left(1+\frac{p}{n}\right)\right]\cos\left[\frac{1}{2}\ln (n^2+pn)\right]\to 0\end{align*}
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.