1
$\begingroup$

$x_n=\frac{n+sin(n)}{n+7}$

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

$\epsilon>|x_{2n}-x_n|=|\frac{sin(n)(2(n+7)Cos(n)-2n-7)+7n}{(2n+7)(n+7)}|\leq\frac{9n+7}{(2n+7)(n+7)}<\frac{9n}{2n^{2}}=\frac{9}{2n}$
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?

$\endgroup$
  • $\begingroup$ 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
2
$\begingroup$

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 $$ |x_{n+k}-x_n|\le\frac8{n+7} $$

| cite | improve this answer | |
$\endgroup$
  • $\begingroup$ 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

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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