0
$\begingroup$

Suppose that {s$_n$} is a convergent sequence. Prove that $$\lim_{n \to \infty} 2s_n $$ exists. I considered countable sets and bijections between natural numbers but I don't know if they apply here. Please help!

$\endgroup$
4
  • $\begingroup$ Do you know what the definition of convergence is? $\endgroup$ Feb 26 '14 at 21:25
  • 1
    $\begingroup$ Try to use the limit laws. Since $(s_n)\rightarrow s$ to say something, then $(2s_n)\rightarrow 2s$ $\endgroup$ Feb 26 '14 at 21:27
  • $\begingroup$ Don't make things too hard on yourself! This is more like calculus than set theory. Think about distances rather than cardinalities. $\endgroup$
    – Alexander Gruber
    Mar 1 '14 at 2:35
  • 1
    $\begingroup$ You've posted three very similar questions in rapid succession. Ask one, and try to understand the answer. Then if you still have problems, ask another question and state specifically what it is you don't understand. $\endgroup$
    – robjohn
    Mar 2 '14 at 1:12
2
$\begingroup$

Convergence to a limit $s$ means that for any $\epsilon > 0$, you can find $N$ so that $n \geq N$ implies $|s_n - s| < \epsilon$. If you consider the sequence $2s_n$, then just choose $N$ so that $n \geq N$ implies $|s_n - s| < \epsilon / 2$ and then you will have that $|2s_n - 2s| < \epsilon$ so your sequence converges to $2s$.

$\endgroup$

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.