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I was wondering if it is possible for a sequence to have a strictly increasing subsequence and a subsequence that converges to 0. I think there is such a sequence and an example would be $a_n$ = $n$/($n$ + $1$). Am I correct?

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    $\begingroup$ Again, nothing difficult here: a_n = -1/n $\endgroup$ – fretty Sep 25 '13 at 22:55
  • $\begingroup$ Your sequence does not converve to $0$, more importantly it converges to $1$. Therefore it can't have subsequences converging to $0$. $\endgroup$ – Git Gud Sep 25 '13 at 22:56
  • $\begingroup$ Just make two sequences, and combine them together somehow. $\endgroup$ – Tunococ Sep 25 '13 at 23:00
  • $\begingroup$ @fretty It's customary, when addressing someone other than the OP, to precede the message with $@\text{ name}$, like I did here. $\endgroup$ – Git Gud Sep 25 '13 at 23:00
  • $\begingroup$ @fretty Isn't $a_n$ = $-1$/$n$ a decreasing sequence? $\endgroup$ – user87274 Sep 25 '13 at 23:01
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Let

$$a_n=\left\{\begin{array}{rcl} 0 & \mbox{if} &n \text{ is odd}, \\ 2n & \mbox{if} & n \text{ is even}. \end{array} \right.$$

The formula for this sequence is $a_n=n+(-1)^n n$.

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  • $\begingroup$ I like the $a_n$ = $-1/n$ example though. $\endgroup$ – user87274 Sep 25 '13 at 23:09

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