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.


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



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 $N\leq n$

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




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

  • $\begingroup$ 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

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 $$ |a_n+b_n - (a+b)| \le |a_n-a|+|b_n-b| < \epsilon/2 + \epsilon/2 = \epsilon.$$


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.

  • $\begingroup$ 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

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