If a monotone sequence is convergent, does this imply boundedness? I was proving a problem which states that: let $\{x_n\}_n$ be a monotone sequence of real numbers. Show that $\{x_n\}_n$ is convergent if and only if it is bounded.
I have proven that if the sequence is monotone and bounded, it is convergent. But must I prove the reverse case where I assume that the sequence is monotone and convergent and prove boundedness? I assume this is trivial because if the sequence is increasing or decreasing and is bounded, then the limit must be the supremum or the infimum.
 A: It's trivial, but for a different reason: a convergent sequence in any metric space is bounded. Suppose $\{x_n\}$ isn't bounded. Then no open ball centered at $x$ contains all points of the sequence. So for each $k$, we can find a point $x_{n_k}$ such that $d(x,x_{n_k})>k$. Take $\varepsilon=1$. Then for any $N$, we have a positive integer $n_k$ such that $d(x,x_{n_k})>k>1$. Hence, $x_n$ does not converge to $x$, a contradiction.
In fact, any Cauchy sequence is bounded. It's a good exercise to prove that as well.
A: Needless to use a proof by contradiction: if $(x_n)$ converges to $\ell$ in a metric space, then for every $\varepsilon>0$ there exists $N$ such that $d(x_n,\ell)<\varepsilon$ for all $n>N$.
Here every $x_n$ belongs to the ball $$B(\ell, \max(\varepsilon, d(x_0,\ell),d(x_1,\ell),\dots, d(x_N,\ell)).$$
A: First, assume the sequence is monotone increasing i.e. $x_n \le x_{n+1}$ for all $n$.
What does it mean that the sequence converges?  It means: there is some $x \in \mathbb{R}$ such that given $\varepsilon > 0$ there is an $N \in \mathbb{N}$ such that if $n \ge N$ then
$$ |x_n - x| < \varepsilon $$
Hence, we have $x_n - x < \varepsilon$ so that $x_n < x + \varepsilon$ for all $n \ge N$.
Now, $x_n$ is monotone, so what can you conclude about $x_n$ for $n < N$?
To wrap everything up, just pick $\varepsilon = 1$.  Then there is an $N$ as above, and so for all $n \ge N$, $x_n < x + 1$.  Hence, the sequence is bounded by $x + 1$.
A: More generally,
a convergent series is bounded;
monotone is not needed.
Proof:
Suppose
$(x_n)_{n=1}^{\infty}$
is convergent
to limit $L$.
Then,
for any $\epsilon > 0$,
there is a $N(\epsilon)$
such that,
for $n > N(\epsilon)$,
$|x_n-L| \le \epsilon$.
Therefore,
for $n > N(\epsilon)$,
$L-\epsilon \le x_n \le L+ \epsilon$.
Now, let
$u(\epsilon)
=\min(x_n)_{n=1}^{N(\epsilon)}
$
and
$v(\epsilon)
=\max(x_n)_{n=1}^{N(\epsilon)}
$.
Then,
for all $n$,
$\min(u(\epsilon), L-\epsilon)
 \le x_n 
\le \max(v(\epsilon), L+ \epsilon)
$,
so the sequence is bounded.
