Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

So sum law of limits tell us

$\lim_{n\to\infty} (a_n+b_n)=X + Y$ if $\lim_{n\to\infty} a_n = X$ and $\lim_{n\to\infty} b_n = Y$

Here is my attempt to prove it.

Proof

Let $\frac{\epsilon}{2}>0$, then $\exists N_a,N_b:$

(1) $|a_n - X| < \frac{\epsilon}{2}$ whenever $n> N_a$

(2) $|b_n - Y| < \frac{\epsilon}{2}$ whenever $n> N_b$

Add (1) and (2) to get $|a_n - X| + |b_n-Y| < \epsilon \implies |a_n + b_n - (X + Y)| \leq |a_n - X| + |b_n-Y| < \epsilon \iff |a_n + b_n - (X + Y)| < \epsilon$

By setting $N=max\left \{ N_a,N_b \right \}$, we get the result.

I consulted an analysis book and they kinda did the same thing, but I think I worked backwards by adding (1) and (2).

share|improve this question

2 Answers 2

up vote 3 down vote accepted

Let $\epsilon>0$. Then there exist positive integers $N_a$ and $N_b$ such that $|a_n-X|<\frac{\epsilon}{2}$ when $n>N_a$ and $|b_n-Y|<\frac{\epsilon}{2}$ when $n>N_b$. Let $N$ be the maximum of $N_a$ and $N_b$. Then for all $n>N$, we have $|a_n+b_n-(X+Y)|\leq|a_n-X|+|b_n-Y|<\frac{\epsilon}{2}+\frac{\epsilon}{2}=\epsilon$.

share|improve this answer
    
why do we (or I guess you lol) start with $|a_n+b_n-(X+Y)|$ first? Why can't I start with $|a_n-X|+|b_n-Y|$ first? –  Hawk Sep 23 '12 at 19:49
1  
Becausew your goal is to prove that $|a_n+b_n-L|$ can be as small as you want, where $L$ is the limit of the sum. Here you take $L=X+Y$ because you "feel" it should be this. –  Jean-Sébastien Sep 23 '12 at 19:53
    
Because inequality $|a_n + b_n - (X + Y)| < |a_n - X| + |b_n-Y|$ not always hold –  M. Strochyk Sep 23 '12 at 19:57
    
@M.Strochyk, a typo on my part. Would it be alright then if I replace < with <= in my proof? I'll do it now –  Hawk Sep 23 '12 at 19:58
1  
It always bugs me when, in these types of proof, people use magic constants with $\epsilon$ ($\epsilon/2$ here) so the result comes out bounded by $\epsilon$. Just use $\epsilon$ in the analysis, and if the bound comes out $2\epsilon$ or $10 \epsilon$ or whatever, as long as it is a finite number times $\epsilon$, that is good enough. –  marty cohen Sep 23 '12 at 20:37

By the triangle inequality $|a_n + b_n - X - Y| \leqslant |a_n - X|+| b_n - Y| \leqslant \frac{\varepsilon}{2}+\frac{\varepsilon}{2}=\varepsilon$ $\forall n\geqslant N=\max\left \{ N_a,N_b \right \}$

share|improve this answer

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.