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.

  • 1
    $\begingroup$ Sounds correct to me, I'd explain why $b_n$ is monotonically increasing but except for that seems completely fine. $\endgroup$ – Yuval Gat Nov 11 '18 at 15:16
  • $\begingroup$ @YuvalGat, thanks, I will add it to my proof. $\endgroup$ – Galc127 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$$a_n=\begin{cases}1&\text{ if }n=1\\0&\text{ otherwise.}\end{cases}$$

  • $\begingroup$ Simple is in the eye of the beholder... Thanks. $\endgroup$ – Galc127 Nov 11 '18 at 15:20

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