3
$\begingroup$

Assume $(a_n)$ and $(b_n)$ are two real sequences, and define $$ c_n = \text{max}\left\{a_n, b_n\right\} $$ for $n \in \mathbb{N}$. Suppose $(a_n)$ and $(b_n)$ are two convergent sequences. Prove then that $(c_n)$ is also a convergent sequence, and that $$ \lim_{n \to \infty} c_n = \text{max} \left\{ \lim_{n \to \infty} a_n, \lim_{n \to \infty} b_n \right\}. $$

Attempt: $(a_n)$ and $(b_n)$ are both convergent. Hence there exists an $n_0 \in \mathbb{N}$ such that $\forall n \geq n_0: | a_n - L | < \epsilon$. Furthermore, there exists an $n_1 \in \mathbb{N}$ such that $\forall n \geq n_1: | b_n - K | < \epsilon $. Here $L$ and $K$ are the limits of resp. $(a_n)$ and $(b_n)$. Now let $n_2 = \text{max} \left\{n_0, n_1 \right\}$. Let $n \geq n_2$ be arbitrary. Then we have that $L - \epsilon < a_n < L + \epsilon$ and $K - \epsilon < b_n < K + \epsilon$. Since $c_n = \text{max}\left\{a_n, b_n\right\}$, we have that $c_n \geq a_n$ and $c_n \geq b_n$. So $ L - \epsilon < a_n \leq c_n$ and thus $ L - \epsilon < c_n$.

But now I'm stuck. Help would be appreciated!

$\endgroup$
1
  • $\begingroup$ I would probably split into three cases: $L<K, L = K$ and $L > K$, and tackle each separately. $\endgroup$ Commented Jan 26, 2016 at 20:38

2 Answers 2

Reset to default
5
$\begingroup$

Remember that $\max\{x,y\}=\frac{x+y+|x-y|}{2}$.

Thus $c_n=\frac{a_n+b_n+|a_n-b_n|}{2}$. So, if $\lim a_n=a$ and $\lim b_n=b$:

$$\lim c_n=\frac{1}{2}(\lim a_n+\lim b_n+|\lim a_n-\lim b_n|)=\frac{1}{2}(a+b+|a-b|)=\max\{a,b\}$$

$\endgroup$
7
  • $\begingroup$ I see, thanks. But what about proving convergence? Should I split that into three cases, depending on the two limits of $(a_n)$ and $(b_n)$? $\endgroup$ Commented Jan 26, 2016 at 20:50
  • $\begingroup$ No!. In general, if $\{x_n\}$ is a convergent sequence and $f:\mathbb{R}\to\mathbb{R}$ is continuos, then $\{f(a_n)\}$ is convergent and $\lim f(a_n)=f(\lim a_n)$. My answer shows that $\{c_n\}$ is convergent and find its limit $\endgroup$ Commented Jan 26, 2016 at 20:52
  • $\begingroup$ @sinbadh You wrote max$(x,y)= \frac{x+y + |x-y|}{2}$. I didn't get the logic behind this formulation. Can you provide me with any source or reasoning behind it? $\endgroup$ Commented Aug 21, 2018 at 19:48
  • $\begingroup$ @Damn1o1 If $x \geq y$, then $x-y$ is positive (or zero) and hence $|x-y|=x-y$. Then the formula simplifies to $\frac{x+y+x-y}{2}=x$. $\endgroup$ Commented Aug 21, 2018 at 20:47
  • $\begingroup$ @JannikPitt What about min$(x,y)$ will it be convergent? $\endgroup$ Commented Aug 22, 2018 at 7:03
2
$\begingroup$

Let $M= \max (L,K).$ When you reach $$n>n_2\implies (\;(L-e<a_n<L+e)\land (K-e<b_n<K+e)\;)$$ you have, for any $n>n_2,$ $$M-e=\max (L,K)-e =\max (L-e,K-e)<\max (a_n,b_n)<\max (L+e,K+e)=$$ $$=\max (L,K)+e=M+e.$$

$\endgroup$
1
  • $\begingroup$ Thank you! The kind of proof I was looking for. $\endgroup$ Commented Jan 26, 2016 at 21:43

You must log in to answer this question.

Start asking to get answers

Find the answer to your question by asking.

Ask question

Explore related questions

See similar questions with these tags.