Determine whether a sequence is bounded above If I want to determine whether a sequence, ${a_n}$, is bounded above $\forall n \in \Bbb{N} $, is it enough to find a sequence that is larger than $a_n$, and show that it converges and is therefore bounded? For example:
$\forall n \in \Bbb{N}, let,$
$$
a_n = \frac{1}{n+1} + \frac{1}{n+2} + ...+\frac{1}{2n}\\
\frac{1}{n+1} + \frac{1}{n+2} + ...+\frac{1}{2n} \leq \frac{1}{n} + \frac{1}{n} + ...+\frac{1}{n} = n\cdot\frac{1}{n}=1
$$
and since $\lim\limits_{n\to\infty}1 = 1$, then 1 must be an upper bound for $a_n$. Is this correct? Thanks.
 A: I think you mess up some ideas.
You say "and since $\lim\limits_{n\to\infty}1 = 1$", but you never showed that $\lim\limits_{n\to\infty}1 = 1$. And if you check the comment of Henry this seems to be wrong. But you don't need the limes.
You showed that
$$a_n = \frac{1}{n+1} + \frac{1}{n+2} + ...+\frac{1}{2n}\\
\frac{1}{n+1} + \frac{1}{n+2} + ...+\frac{1}{2n} \leq \frac{1}{n} + \frac{1}{n} + ...+\frac{1}{n} = n\cdot\frac{1}{n}=1$$
this means
$$a_n\le 1,\; \forall n \in \Bbb{N}$$
And this means that $a_n$ is bounded above by $1$. There is nothing else to show.
Remark 1:
An increasing sequence that is bounded above is convergent
We have
$$a_{n+1}=a_n+\frac{1}{(2n+1)(2n+2)}$$
This means $$a_{n+1}>a_n$$ and so $a_n$ is monotone increasing. If a sequence is increasing and bounded above then the sequence is convergent.
Remark 2:
An convergent sequence is bounded
If a sequence $a_n$ converges to $a$ then there exists a number $N$ such that 
$$\mid a_n-a\mid\le 1,\; \forall n>N$$
and so we have $$a_n\le a+1,\; \forall n>N$$ and
$$a_n\le\max\{a_1,\ldots,a_N\},\; \forall n\le N$$
and therefore the sequence $a_n$ is bounded by $$\max\{N,a_1,\ldots,a_N\}$$
A: Yes, it is correct, we have $a_n \le 1$ for all $n$.
A: Yes, this is enough. If $(b_n)$ dominates $(a_n)$ and $b_n \rightarrow b$, then there is $S\in \mathbb R$ with $b_n<S$ for all $n \in \mathbb N$. Since $(b_n)$ dominates, we get $S>b_n \geq a_n$, so $(a_n)$ is bounded as well.
