I'm still new on this field and I would be so pleased,if someone let me know that my steps are valid and I haven't done any mistakes

My attemps for showing that $|x_{n+1}-x_n| < \epsilon$ failed because I'm having 2 terms $sin(n)$ and $sin(n+1)$, and both of them cannot get their max values for same $n$.

And when I chose $n'$ as $2n$, I've got

So now I can assume that $n>\frac{9}{2\epsilon}$ and choosing $n_0=[\frac{9}{2\epsilon}]+1$ this statement would remain true.
$\forall\epsilon>0,\exists n_0=[\frac{9}{2\epsilon}]+1, \forall n,n' \in N, |x_{n'}-x_n|<\epsilon$ so I can assume that this is Cauchy sequence so is convergent.I'm not really sure that this is true for $n_0$ I've got. Also 1 more question, Is there more simple way to prove this?

  • $\begingroup$ well, the goal is not getting the limit, but to prove convergence by Cauchy's criterion. I'm thinking there might be a more simple way, than I used $\endgroup$ – shcolf Jan 7 '15 at 15:54

Note that for all $n\ge1$, $$ \frac{n-1}{n+7}\le\frac{n+\sin(n)}{n+7}\le\frac{n+1}{n+7} $$ Thus, $$ \frac{n+k-1}{n+k+7}-\frac{n+1}{n+7}\le x_{n+k}-x_n\le\frac{n+k+1}{n+k+7}-\frac{n-1}{n+7} $$ which can be written as $$ \frac6{n+7}-\frac8{n+k+7}\le x_{n+k}-x_n\le\frac8{n+7}-\frac6{n+k+7} $$ The last inequality implies that $$ |x_{n+k}-x_n|\le\frac8{n+7} $$

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  • $\begingroup$ This statement is true enough. but how to get $n(\epsilon)$ dependence from it to get starting point for a formal proof, as you have also added another value k. $\endgroup$ – shcolf Jan 7 '15 at 16:25
  • $\begingroup$ I've added more. Can you see how to get what you want now? $\endgroup$ – robjohn Jan 7 '15 at 16:46

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