Does there exists this kind of real sequence? Does the sequence $(a_n)$ with the following properties exists? 


*

*$(a_n)$ converges to $a$.

*$(a_n)$ is strictly increasing.

*$(a_n)$ is not bounded by $\max\{a_1, a\}$.


Also, I need to find a sequence $(b_n)$ such that $(\frac {1}{n})$ and $(\frac {-1}{n})$ are subsequences of $(b_n)$, but $(b_n)$ is unbounded.
I have been trying to find examples, but nothing useful  comes to my mind. Any ideas ?
Thanks in advance
Cheers
 A: If $(a_n)$ is strictly increasing with limit $a$, then you have $a_1\leq a_n<a$, no matter what. Obviously, this sequence is bounded by $\max\{|a_1|, |a|\}$.
If this is what you mean by $\max\{a_1,a\}$, then the answer is NO, such sequence does not exist. If you mean something different by that, like that $|a_n|\leq\max\{a_1,a\}$, then a sequence $a_1=-100$ and $a_n=1-1/n$ for $n\geq2$ seems to be what you are looking for.
We can actually go further and take the sequence $a_n=-1/n$, it's increasing and converging to $a=0$, and no element satistfies $|a_n|<\max\{a_n,a\}=0$.

For your 2nd question, take any shuffle (interleave) of the sequences $(1/n)$, $(-1/n)$ and $(n)$ or any other unbounded sequence, and it will satisfy your requirements.
A: If a function is strictly increasing and converges, then its limit is necessarily an upper bound.
A: First question: no. If $(a_n)$ is strictly increasing, and converges to $a$, then $\forall n$, $(a_n)< a$. Which you can show by "Reductio ad absurdum". Since $a_1 < a$, $\max\{a_1,a\} =a$ (strange condition), $(a_n)$ is boounded by $a$, no question.
Second question: write it modulo $3$: $k = 3n+m$, $0\le m \le 2$. Then define, for  $k> 2$:


*

*$b_k = {n}$ if $m = 0$, 

*$b_k = \frac{1}{n}$ if $m = 1$, 

*$b_k = -\frac{1}{n}$ if $m = 2$,

*and if you want $b_0 = b_1=b_2 = 0$ to avoid troubles at the beginning.

