Show that $\lim _{n \to \infty}\left (a_{n} + b_{n}\right) = a + b$ help ??

Question

Assume that $\lim_{n \to \infty} a_n=a$ and $\lim_{n \to \infty} b_n=b$

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

2. and that $\lim_{n \to \infty} a_nb_n=ab$ by using the definition of convergence of sequences (well this is two seperate questions, but I guess the principles are pretty much the same)

I mean can I just say that the limit of a sequence is equivavent to the limit of a function as the limit approaches infinity? done? or do I actually have to "prove" it? if yes then I have no idea how to do that :D thanks for tips/solutions/advice :D

Answer 1

We have that $\lim_{n \to \infty} a_n = a$ and $\lim_{n \to \infty} b_n = b$

So $\forall \epsilon > 0 $ $\exists N_{1} \in \mathbb{N}$ such that $|a_{n} - a| < \epsilon/2$ $\forall n>N_{1}$

And also $\exists N_{2} \in \mathbb{N}$ such that $|b_{n} - b| < \epsilon/2$ $\forall n>N_{2}$

Now, let $N = max(N_{1}, N_{2})$ then $\forall n>N$ we have $|(a_{n} + b_{n}) - (a + b)| \leq |a_{n} -a| + |b_{n} - b|$ by the triangle law and then we immediately have $|(a_{n} + b_{n}) - (a + b)| \leq \epsilon /2 + \epsilon /2 = \epsilon$.

So $\lim_{n \to \infty} a_n + b_n = a + b$.

Now, for the product we have to be a little bit sneakier. As $a_n$ tends to a limit $a$, we can show that the sequence is itself bounded: let $\epsilon > 0$ then $\exists N \in \mathbb{N}$ such that $|a_n - a| < \epsilon $ $\forall n > N$. Then we have $|a_n| < |a| + \epsilon$ $ $ $\forall n>N$. Now let $h =$ max $(|a_1|,|a_2|,...,|a_N|, |a| + \epsilon) $ then $|a_n| \leq h$ $\forall n$.

Now $\exists N_1 \in \mathbb{N}$ such that $|a_n - a| < \frac {\epsilon} {2(1 + |b|)}$ and $\exists N_2 \in \mathbb{N}$ such that $|b_n - b| < \frac {\epsilon} {2(1 + h)}$ where h is the maximum of $a_n$ defined in the last paragraph. W use the "$1 + ..$" in the denominator to avoid the awkward cases where either $b$ or $h$ is $0$

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This is a really standard trick: we let $N = $max$(N_1, N_2)$, then we have $|a_n b_n - ab| = |a_n b_n - b a_n + b a_n - ab|$ (this is also a really standard trick, adding and subtracting the same thing from a modulus so that we can use the triangle law, which is what we do now). So we get:

$$ |a_n b_n - ab| = |a_n b_n - b a_n + b a_n - ab| \leq |a_n b_n - b a_n| + |b a_n -ab| = |a_n| |b_n - b| + |b||a_n - a| \leq \frac {\epsilon |a_n| } {2(1 + h)} + \frac {\epsilon b } {2(1 + b)} \leq \epsilon /2 + \epsilon /2 = \epsilon $$

And so we get $\lim_{n \to \infty} a_n b_n = ab$.

As an exercise for yourself you might like to try proving the quotient rule, that is if $b \neq 0$ then $lim_{n \to \infty} \frac {a_n} {b_n} = \frac {a} {b}$ .

Answer 2

Some guidance: For these kind of questions I always find it very useful to write down what we know. So in this case:

Since $\{a_n\}_{n=1}^\infty$ is convergent, by definition, there exists a $N_1 \in \mathbb{N}$ such that for all $\varepsilon > 0$ and for all $n \in \mathbb{N}$ with $n \geq N$ implies $|a_n - a| < \dfrac{\varepsilon}{2}$.

You can do the same for $\{b_n\}_{n=1}^\infty$.

Then, you have to find a suitable $N \in \mathbb{N}$.

After that, you should apply T. Bonger's hint. And then finally, reach the desired conclusion!

For the second question, do the same but note that $|a_nb_n - ab| = |a_nb_n - a_nb + a_nb - ab|\leq |a_nb_n - a_nb| + |a_nb - ab| = |a_n||b_n - b| + |b||a_n - a|$

Answer 3

Suppose $a_n \to a$ and $b_n \to b$.

1) Since $a_n -a \to 0$ and $b_n -b \to 0$ why? and $a_n +b_n -(a+b)=(a_n-a)+(b_n -b)$, it will suffice to show that if $c_n \to 0$ and $d_n \to 0$ so $c_n+d_n \to 0 $. Given $\varepsilon>0$ let choose $N$ such that $|c_n|< \varepsilon/2$ and $|d_n|< \varepsilon/2$ for all $n\ge N$ at the same time. Hence

$$|c_n +d_n|\le|c_n|+|d_n|< \varepsilon$$

as desired.

2) We proof two special cases if $c_n \to c$ and $k\in \mathbb{R}$ so $kc_n \to kc$. Choose $n_0$ such that $|c_n-c|<\varepsilon/(|k|+1)$ for all $n\ge n_0$. So $$|kc_n-kc|=|k||c_n-c|< \varepsilon$$

and the second special case. If $c_n \to c$ and $d_n \to d$, so $(c_n-c)(d_n-d)\to 0$. Given $\varepsilon>0$. Choose $n_0$ such that $|c_n-c|< \sqrt{\varepsilon}$ and $|d_n-d|<\sqrt{\varepsilon}$. Hence $$|(c_n-c)(d_n-d)|=|c_n-c||d_n-d|< \varepsilon$$

For the general case using $(1)$ and the special cases:

\begin{align}\lim_n (a_nb_n-ab)=\lim_n\big[(a_n-a)(b_n-b)+b(a_n-a)+a(b_n-b)\big]\\ =\lim_n(a_n-a)(b_n-b)+b\lim_n(a_n-a)+a\lim_n(b_n-b)\\ =0+b\cdot 0+a\cdot 0=0\end{align}