Answer to Question #176989 in Real Analysis for Ravi

Question

Give an example of a divergent sequence which has two convergent sequences. Justify  your claim

Answer 1

$\displaystyle \textsf{Let}\,\, a_n = \{(-1)^n\}_{n = 1}^{\infty} = \{-1, 1, -1, 1, -1, 1,\ldots\} \\ \textsf{Let}\,\, a_{n1}\,\, \textsf{and}\,\, a_{n2}\,\, \textsf{be subsequences of}\,\, a_n \\ \textsf{Let}\,\,\, a_{n1} = \{-1,-1,-1,\ldots\} \\ a_{n1} \,\,\, \textsf{converges to}\,\, -1 \\ \textsf{Let}\,\,\, a_{n2} = \{1,1,1,\ldots\} \\ a_{n2} \,\,\, \textsf{converges to}\,\, 1 \\ \textbf{Proof that}\,\,\textbf{a}_n \,\, \textbf{diverges} \\ \textsf{Assume that}\,\, a_n\,\, \textsf{converges to}\,\, a^*\,\, \textsf{for any}\,\, \epsilon > 0.\\ |a_n - a^*| < \epsilon\\ \textsf{Let}\,\, \epsilon = \frac{1}{2} \\ |(-1)^n - a^*| < \frac{1}{2}\\ |(-1)^{n + 1} - (-1)^{n + 2}| = 2 \\ |(-1)^{n + 1} - a^* + a^* - (-1)^{n + 2}| = 2 \\ \begin{aligned} 2 &= |(-1)^{n + 1} - a^* + a^* - (-1)^{n + 2}| \\&\leq |(-1)^{n + 1} - a^*| + |a^* - (-1)^{n + 2}| \\ &\leq |(-1)^{n + 1} - a^*| + |(-1)^{n + 2} - a^*| < \frac{1}{2} + \frac{1}{2} = 1 \end{aligned}\\ \textsf{This implies that}\,\, 2 <1, \,\, \textsf{which is false}.\\ \textsf{Hence the sequence is divergent.}$