$\max\{a_1,a_2,\dots,a_n\}$ 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.

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