1
$\begingroup$

Hey so I'm doing some revision and I'm stuck on this question, It's a pretty basic question so I'm not sure why I'm stuck, I could be missing something or overthinking it.

Question

Assume $\lim_{n\to\infty}(a_n)=a, \lim_{n\to\infty}(b_n)=b$ then $\lim_{n\to\infty}(a_n+b_n)=a+b$

Working

$\lim_{n\to\infty}(a_n+b_n)=a+b=\lim_{n\to\infty}(a_n)+\lim_{n\to\infty}(b_n)$

Fix $\epsilon>0$ then $\exists N_1\in\mathbb{N} \space: N_1\leq n$

$\implies |a_n-a|<\frac{\epsilon}{2}$

$\exists N_2\in\mathbb{N} \space : N_2\leq n$

$\implies |b_n-b|<\frac{\epsilon}{2}$

Take $N=$ $\max${$N_1, N_2$} where $N\leq n$

$\therefore |a_n+b_n-a-b|<\epsilon$

$|a_n-a+b_n-b|<\epsilon$

$|a_n-a|+|b_n-b|<\epsilon$

Note

This just doesn't seem right or complete as $\epsilon \nless \epsilon$, so any help would greatly be appreciated!! :)

$\endgroup$
  • $\begingroup$ The last inequality is not necessary. $\endgroup$ – IAmNoOne Apr 26 '17 at 3:08
  • $\begingroup$ You almost got there, $N = \max(N_1, N_2) $, let $n \le N$, $: | a_n +b_n - (a + b) | = | (a_n- a) + (b_n -b) | \le |a_n -a| + |b_n -b| \lt \epsilon /2 + \epsilon /2 = \epsilon$. $\endgroup$ – Peter Szilas Apr 26 '17 at 4:35
2
$\begingroup$

You are not supposed to assume $|a_n+b_n - (a+b)| < \epsilon,$ you are supposed show it is true when $n>N$.

You picked the right $N,$ ($\max(N_1,N_2))$ where $N_1$ and $N_2$ are picked to have the properties you wrote down (and whose existences are guaranteed by the fact that $\lim a_n =a$ and $\lim b_n = b$).

Then, the proper way to show that $|a_n+b_n - (a+b)| < \epsilon$ for all $n>N$ is through the chain of inequalities $$ |a_n+b_n - (a+b)| \le |a_n-a|+|b_n-b| < \epsilon/2 + \epsilon/2 = \epsilon.$$

$\endgroup$
-1
$\begingroup$

After you have taken $N$ to be the larger of the $\{N_1,N_2\}$ you can write:

$$|a_n-a+b_n-b|\leq |a_n-a|+|b_n-b|<\frac{\epsilon}{2}+\frac{\epsilon}{2} = \epsilon, \quad (n>N)$$

Which is your desired inequality.

$\endgroup$
  • $\begingroup$ Why the downvote? $\endgroup$ – Bernard W Apr 26 '17 at 3:50
  • $\begingroup$ @Bernhard. Do not get it either, the downvote. You pinpointed where James should have followed through. What else is there to say. $\endgroup$ – Peter Szilas Apr 26 '17 at 8:00

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.