2
$\begingroup$

Show that if $x_n=O(\alpha_n)$, then $cx_n=O(\alpha_n)$

I know that $x_n=O(α_n)$ means there exists a constant $C$ and integer $n_0$ such that $|x_n|≤C|α_n|$ for all $n≥n_0$ with which $|c||x_n|≤C|c||α_n|$ and so $|cx_n|≤C|c||α_n|$, making $K:=C|c|$ we get that $|cx_n|≤K|α_n|$ for all $n≥n_0$, then we have the result. Is this reasoning right? Thank you very much.

$\endgroup$
  • 4
    $\begingroup$ The reasoning is perfectly valid. $\endgroup$ – Simply Beautiful Art Dec 14 '17 at 2:52

Your Answer

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

Browse other questions tagged or ask your own question.