# Help in proving $\lim \limits_{n \to \infty} (a_n \pm b_n) = a \pm b$

This was left as a non-assessed exercise, and I am unsure of how to attack it:

If $a_n$ and $b_n$ are convergent sequences converging to $a$ and $b$ respectively. Then prove that:

## $$\lim \limits_{n \to \infty} (a_n \pm b_n) = a \pm b$$

I am thinking I will have to call forth content in regards to $\epsilon - N$. But I have no $n$ to work with, so I have no idea where to take this. Any tips or advice is appreciated!

Note: I have seen Prove for all sequences $\{a_n\}$ and $\{b_n\}$, if $\lim a_n = a$ and $\lim b_n = b$, then $\lim a_n + b_n = a+b$ entirely in first-order logic, but I can't see this as being an efficient way of proving this, nor can I follow it entirely.

## 3 Answers

$\lim \limits_{n \to \infty} a_n=a$ means that $|a_n-a|\to 0$ similarly $\lim \limits_{n \to \infty} b_n=b$ means that $|b_n-b|\to 0$. Therefore $|a_n\pm b_n-(a\pm b)|\leq |a_n-a|+|b_n-b|\to 0$

Of course, if you don't know that $\lim \limits_{n \to \infty}a_n\to a$ and $\lim \limits_{n \to \infty}b_n\to b$ the claim makes no sense.

• Great answer. That makes perfect sense! – Display Name Apr 3 '14 at 23:48

What you need to do is to consider $a_n$ and $b_n$ separately. You can get as close as you like to $a$ and $b$ respectively, so you can get within $\frac {\epsilon}2$ with $N_a, N_b$ respectively.

Now you take an $N$ at least as big as these, and you have $\frac {\epsilon}2$ simultaneously for both. Use the "triangle inequality" (inherent in the metric) to prove that you are within $\epsilon$ for the sum and difference.

I presume you have $a_{n} \to a$, $b_{n} \to b$ in your hypothesis. By definition of limit you have:

$\forall \delta > 0 \quad \exists N \in \mathbb{N} \quad s.t. |a_{n} - a| \leq \delta \quad \forall n \geq N$

and of course you have the analogous for $b_{n}$. For clearness, call the first $N=N_{a,\delta}$ and the second $N_{b,\delta}$. Then you would like to fix $\varepsilon >0$ and write the property of limit for the sum for this fixed $\varepsilon$. Then take $N = max\left( N_{a,\frac{\varepsilon}{2}},N_{a,\frac{\varepsilon}{2}}\right)$, then you have

$|a_{n}+b_{n}-a-b| \leq |a_{n}-a| + |b_{n}-b|\leq \varepsilon \quad \forall n \geq N$.

Analogously for $a_{n}-b_{n}$.