does the sequence ${a_n}$ converges implies it is bounded? does the sequence ${a_n}$ converges implies it is bounded?
i think the answer is no, but i've seen a proof used this fact and our TA used this fact in class as the following question:
arithmetic mean of a sequence converges
 A: Yes:
Suppose the sequence converves to some $L\in\mathbb R$. Then by definition of convergence, there is an $N$ such that all terms in the squence after $a_n$ are within a distance of $1$ from $L$.
If all terms have a distance to $L$ of at most $1$, then we're done, of course. Otherwise, there are at most $N$ elements of the sequence whose distance to $L$ is more than $1$. That is finitely many, so one of them will be farthest from $L$. Let the distance from that term to $L$ be $a$.
Now every term in the sequence is within the open ball $B_{a+1}(L)$. Thus, it is bounded.
A: The answer would be YES. (I assume that your setting is in the real number)
Now Assume it converges to a finite number a, then for large N, $a_n$ would be close enough to your limit a for all $n>N$ which means they are uniformly bounded by some number which we shall call it $M$. Finally, pick the maximum of the first N terms $a_1.....a_N$ and this uniform bound $M$, this maximum will be your upper bound for the whole sequence.
A: $a_n$ converges means that
there is a limit $L$
such that,
for any $\epsilon > 0$,
there is an $N(\epsilon)$
such that
$|a_n-L|
< \epsilon$
for $n > N(\epsilon)$.
Set $\epsilon = 1$.
Then
$|a_n - L| < 1$
for $n > N(1)$.
Therefore
$L-1 < a_n < L+1$
for $n > N(1)$,
so that the $a_n$
are bounded for $n > N(1)$
by $|L|+1$.
Since there are only a finite number of
$n \le N(1)$,
there is a bound on those $a_n$
of $\max_{n=1}^{N(1)} |a_n|$.
Therefore all the $a_n$ are bounded
by
$\max(|L|+1, \max_{n=1}^{N(1)} |a_n|)$.
A: The answer is yes. Since $(a_n)_{n\in\Bbb N}$ is convergent by assumption, let $a$ its limit. Then for every $\varepsilon >0$ there exists $m\in\Bbb N$ such that $|a_n-a|<\varepsilon$ for every $n>m$. In particular there exist $\bar{m}$ such that $|a_n|<1+|a|$ for every $n>\bar{m}$.
Now the set $\{a_1,\cdots a_\bar{m}\}$ is finite, and let $M$ the maximum of their modulus. Now the $\max\{M,1+|a|\}$ gives the right bound.
