Vtyjkyuk

If I want to determine whether a sequence, $a_n$, is bounded above $forall n in BbbN $, is it enough to find a sequence that is larger than $a_n$, and show that it converges and is therefore bounded? For example:

$forall n in BbbN, let,$

$$
a_n = frac1n+1 + frac1n+2 + ...+frac12n\
frac1n+1 + frac1n+2 + ...+frac12n leq frac1n + frac1n + ...+frac1n = ncdotfrac1n=1
$$
and since $limlimits_ntoinfty1 = 1$, then 1 must be an upper bound for $a_n$. Is this correct? Thanks.

I think you mess up some ideas.

You say "and since $limlimits_ntoinfty1 = 1$", but you never showed that $limlimits_ntoinfty1 = 1$. And if you check the comment of Henry this seems to be wrong. But you don't need the limes.

You showed that

$$a_n = frac1n+1 + frac1n+2 + ...+frac12n\
frac1n+1 + frac1n+2 + ...+frac12n leq frac1n + frac1n + ...+frac1n = ncdotfrac1n=1$$
this means

$$a_nle 1,; forall n in BbbN$$

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
$$a_n+1=a_n+frac1(2n+1)(2n+2)$$
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
$$mid a_n-amidle 1,; forall n>N$$
and so we have $$a_nle a+1,; forall n>N$$ and
$$a_nlemaxa_1,ldots,a_N,; forall nle N$$
and therefore the sequence $a_n$ is bounded by $$maxN,a_1,ldots,a_N$$

Yes, it is correct, we have $a_n le 1$ for all $n$.

Yes, this is enough. If $(b_n)$ dominates $(a_n)$ and $b_n rightarrow b$, then there is $Sin mathbb R$ 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.