Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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?

share|improve this question
    
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
add comment

2 Answers

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

share|improve this answer
add comment

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.

share|improve this answer
add comment

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

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