Suppose that ($S_n$) converges and has only finitely many distinct terms . Show that $S_n$ is constant for large $n$
What I have tried
Let $M$ the set of distinct elements of the sequence ($S_n$) $M$={$t_1,t_2,….,t_r$} let $m$=min{$|t_i-t_j| $: $i$ is not equal to $j$} due to ($S_n$) is convergent for any positive epsilon there is $N$ such that $|S_n-S|<$ epsilon whenever $n>N$ let epsilon = m so there is $N_m$ Such that $|S_n-S|<$ $m$ whenever $n>N_m$ this satisfies if $|S_n-S|=0$ so $S_n=S$ whenever $n>N_m$
Best Answer
Let $s=\lim_{n\rightarrow\infty} S_n$. Since $S_n$ has only finitely many distinct terms there are infinitely many terms of $S_n$ that have the same value let $t$. Consider the subsequence $\{S_{k_n}\}_n$ that consists of all these terms i.e. $S_{n_k}=t=constant$. Then $$s=\lim_{n\rightarrow\infty} S_n=\lim_{n\rightarrow\infty} S_{k_n}=t$$ since for a convergent sequence all subsequences converge to the same limit.
Now, by the definition of the limit we have that for large $n$ infinitely many terms of $S_n$ lie in an $\varepsilon$-ball around the value $t$ (where we choose some $\varepsilon >0$).
Morever, since $S_n$ attains finite distinct values for $\varepsilon >0$ small enough and large $n$ the $\varepsilon$-ball around the value $t$ doesn't contain any other of the finitely many values of the terms of $S_n$, i.e. for large $n$ the sequence is constant and equal to $t$.