Question
Show that a sequence of real numbers converges if and only if it is bounded and has not more than one accumulation point
Proof:
Let $(a_n)_{n\in\mathbb N}$ be a sequence which converges to $\alpha$. Let $\beta \ne \alpha\space$. Where $\beta\space$ is another accumulation point. Hence there is a subsequence such that $\lim_{k\to\infty}a_{n_k}=\beta$.
For $\epsilon := \frac{\alpha+\beta}{2}\gt0$
$\exists N\in\mathbb N$ such that $|a_{n}-\alpha|\lt\epsilon\space$ $\forall n\gt N$ and $\exists K\in\mathbb N$ such that $|a_{n_k}-\beta|\lt\epsilon\space$ $\forall k\gt K$.
Now choose $k^*\in\mathbb N$ such that both $k^* \gt K$ and $n_{k^*}\gt N$:
$|\alpha -\beta|\le|\alpha-a_{n_k^*}|+|a_{n_k^*}-\beta|\lt2\epsilon=|\alpha -\beta|$ which is a contradiction so there is only one accumulation point.
Also, choosing $\epsilon$ to be some positive number;
Let $\epsilon =1$:
$\Rightarrow |a_n-\alpha|\lt 1\Rightarrow |a_n|-|\alpha|\le|a_n-\alpha|\lt 1 \Rightarrow |a_n|\lt |\alpha|+1$
So if $n\gt N$, then $|a_n|\lt 1+|\alpha|$
Now consider where $n\le N$. This is a finite set so there exists a maximum value, call it $∣a_p∣$, that is $\max{(∣a_1∣,∣a_2∣,...,∣a_p∣,...,∣a_N∣})=|a_p∣$.
Let $M=\max({|a_p∣, 1+|\alpha|})$
$\forall n$, $|a_n|\le M$.
Hence $a_n$ is bounded
$\therefore$ Since $a_n$ converges $\Rightarrow$ $a_n$ bounded and has not more than one accumulation point.
Comment
This a question I have been asked in Analysis I. The question asks to prove it in bother directions (if and only if). I am unsure on how to do this in a succinct manner. Any tips/alternative proofs are really appreciated :)
You have correctly proven the easy direction: If $\lim_{n\to\infty} a_n=\alpha$ then the sequence $(a_n)_{n\geq0}$ is bounded and cannot have another accumulation point $\beta\ne\alpha$.
For the other direction we have to consider an arbitrary sequence $n\mapsto a_n\in{\mathbb R}$ which is bounded, i.e., $|a_n|\leq M$ for some $M$, and has at most one accumulation point. In this case it has exactly one accumulation point $\alpha\in [-M,M]$, since $[-M,M]$ is compact. If $\lim_{n\to\infty}a_n=\alpha$ is wrong then there is an $\epsilon_0>0$ such that there are arbitrarily large $n$ with $|a_n-\alpha|\geq\epsilon_0$. These bad $a_n$ would lie in the compact set $S:=[-M,M]\>\setminus\>]\alpha-\epsilon_0,\alpha+\epsilon_0[\>$ and therefore would have an accumulation point $\beta\ne\alpha$, contrary to assumption.