Limit of a Sequence 02

" $\displaystyle{\lim_{n \rightarrow \infty} a_n}$ exists"; This is true, but more can be said. If $\displaystyle{\lim_{n \rightarrow \infty} a_n} =a > 1$ , then there is some $\epsilon > 0$ such that $a- \epsilon \geq 1$ ; i.e. for all $n \geq N$ we have $\left| a_n - a\right| \geq \epsilon$ , meaning $a_n$ does not converge to $a$ . Contradiction.

" $\displaystyle{\lim_{n \rightarrow \infty} a_n}$ exists and $\displaystyle{\lim_{n \rightarrow \infty} a_n} < 1$ "; Let $\displaystyle{a_n = \frac{n}{n+1}}$ . Then $\displaystyle{a_n < 1}$ for all $n \geq 0$ , but $\displaystyle{\lim_{n \rightarrow \infty} a_n = 1}$ .

" $\displaystyle{\lim_{n \rightarrow \infty} a_n}$ exists and $\displaystyle{\lim_{n \rightarrow \infty} a_n} = 1$ "; Let $\displaystyle{a_n = \frac{n}{2(n+1)}}$ . Then $\displaystyle{a_n < 1}$ for all $n \geq 0$ , but $\displaystyle{\lim_{n \rightarrow \infty} a_n = \frac{1}{2}}$ .

"None of these is necessarily true."; If a sequence is increasing and bounded above, then it converges.