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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!

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  • $\begingroup$ I would probably split into three cases: $L<K, L = K$ and $L > K$, and tackle each separately. $\endgroup$ – Arthur Jan 26 '16 at 20:38
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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 $\lim b_n=b$:

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

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  • $\begingroup$ I see, thanks. But what about proving convergence? Should I split that into three cases, depending on the two limits of $(a_n)$ and $(b_n)$? $\endgroup$ – Kamil Jan 26 '16 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 $\endgroup$ – sinbadh Jan 26 '16 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? $\endgroup$ – Ronald Aug 21 '18 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 $\frac{x+y+x-y}{2}=x$. $\endgroup$ – Jannik Pitt Aug 21 '18 at 20:47
  • $\begingroup$ @JannikPitt What about min$(x,y)$ will it be convergent? $\endgroup$ – Ronald Aug 22 '18 at 7:03
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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 $n>n_2,$ $$M-e=\max (L,K)-e =\max (L-e,K-e)<\max (a_n,b_n)<\max (L+e,K+e)=$$ $$=\max (L,K)+e=M+e.$$

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  • $\begingroup$ Thank you! The kind of proof I was looking for. $\endgroup$ – Kamil Jan 26 '16 at 21:43

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