# $\max\{a_1,a_2,\dots,a_n\}$ converges for a convergent sequence $a_n$

I am tackling the following question and want to be sure that my reasoning is fine.

Let $$a_n$$ be a convergent sequence s.t $$\displaystyle \lim_{n\to\infty}a_n=a$$. Let $$b_n\triangleq\max\{a_1,a_2,\dots,a_n\}$$ Prove that $$b_n$$ converges. Also, is it necessarily the case $$\displaystyle \lim_{n\to\infty}b_n=a$$?

My try:

As $$a_n$$ converges, it is bounded. Let $$M$$ be an upper bound of $$a_n$$. We note that it is also an upper bound of $$b_n$$ and that $$b_n$$ is monotonically increasing, thus $$b_n$$ converges.

I looked at the sequence $$a_n=\dfrac{1}{n}$$. We have $$\displaystyle \lim_{n\to\infty}a_n=0$$ and $$\forall n\in\mathbb{N}:\ b_n=1$$, i.e $$\displaystyle \lim_{n\to\infty}b_n=1\ne\lim_{n\to\infty}a_n$$

It seems too simple and I believe that I am missing something.

Any comment regarding the solution will be appreciated. In the case it is wrong I will be thankful for some hints in the right direction. Thanks.

• Sounds correct to me, I'd explain why $b_n$ is monotonically increasing but except for that seems completely fine. Nov 11 '18 at 15:16
• @YuvalGat, thanks, I will add it to my proof. Nov 11 '18 at 15:19

What you did is fine and if you missed something is that an even simpler example than yours can be found. Just take$$a_n=\begin{cases}1&\text{ if }n=1\\0&\text{ otherwise.}\end{cases}$$