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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

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    What is the sequence in? Is it a sequence of reals? Extended reals? – 2017-02-10
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    The only information given about this sequence is $lim_{n\to\infty} a_n = L$, thank you! – 2017-02-10
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    It is true, unless one makes insane conventional choices, like letting the sequence be in the extended reals, $[-\infty,\infty]$, setting $a_1=\infty$, and $a_n=1$ for $n>1$. But then we need to still use the ridiculous convention that infinite quantities in the extended reals are unbounded. – 2017-02-10
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    If $L$ is real the answer is yes, if $L$ is extended real the answer is no. Even if $a_n$ real when $L$ is allowed to be extended real the answer is no as the example $a_n=n$ shows. – 2017-02-10
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    @dioid: I don't think it makes much sense to interpret this question in the extended reals, because that is not a metric space at all (though it is metrizable), and therefore speaking of "bounded" is not even meaningful. – 2017-02-10
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    @Henning Makholm I would agree to that. I just wasn't able to find the justification you just provided. Thanks. – 2017-02-10

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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.

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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.

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$a_n$ 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 $\max(|L|+1, \max_{n=1}^{N(1)} |a_n|)$.

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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.