0
$\begingroup$

Let us consider two sequences $(a_n)$ and $(b_n)$ with equal limits denoted as $g$ for $n \to \infty$. Let us take

$$c_n=\left\{ \begin{array}{ccc} a_n&\mbox{for odd n,}\\ b_n&\mbox{for even n.} \end{array} \right.$$

Using the definition of the limit show that $\lim_{\,n \to \infty} c_n = g$.

Let us take $\epsilon>0.$ Since $\lim_{\,n \to \infty} a_n = g$ and $\lim_{\,n \to \infty} b_n = g$, we have that there is an $N_1 \in \mathbb{N}$ and an $N_2 \in \mathbb{N}$ such that for all $n>N_1$

$$|a_n-g|<\epsilon$$ and for all $n>N_2$ $$|b_n-g|<\epsilon.$$

I do not know how to use this to show that $|c_n-g|$ is also bounded. I would be grateful for any help.

$\endgroup$
  • 1
    $\begingroup$ What can you say about $|c_n-g|$ when $n>\max(N_1,N_2)$? $\endgroup$ – TonyK Feb 10 at 16:18
  • $\begingroup$ It is bounded with $\epsilon$ for odd $n$ (since $|a_n-g|$ is bounded) and it is bounded with $\epsilon$ for even $n$ (since $|b_n-g|$ is bounded), so it is bounded in general. That is it? $\endgroup$ – Uhans Feb 10 at 16:22
  • $\begingroup$ Exactly.${}{}{}$ $\endgroup$ – TonyK Feb 10 at 16:23
  • $\begingroup$ Thank you! Now it is clear. $\endgroup$ – Uhans Feb 10 at 16:27
3
$\begingroup$

Take $N := \max\{N_1,N_2\}$. Then for $n > N$, if $c_n = a_n$ then $|c_n - g| = |a_n - g| < \epsilon$ as $n > N_1$. If $c_n = b_n$ then $|c_n - g| = |b_n - g| < \epsilon$ as $n > N_2$.

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

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.