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True or false: a sequence $\{a_{n}\}$ is convergent if and only if the subsequences $a_{2n}$ and $c_{2n + 1}$ both converge to the same number

I think that the answer is true. I know that the converse is true because the subsequences of a convergent sequence always converge to the same number. I'm not sure about the forward direction though. I couldn't come up with a counterexample.

  • $c_{2n+1}$? Did you mean $a_{2n+1}$? – Alex R. Dec 14 '18 at 00:53
  • The forward direction is true as well. To see this, consider the contrapositive... Suppose that it is not true that $a_{2n}$ and $a_{2n+1}$ both converge to the same number. It could be that is a result of $a_{2n}$ not converging at all for example, in which case $a_n$ doesn't converge, or it could be that $a_{2n}$ converges to a number different than $a_{2n+1}$, in which case again $a_n$ wouldn't converge. – JMoravitz Dec 14 '18 at 00:53
  • yes, i mean $a_{2n + 1}$ –  Dec 14 '18 at 00:56
  • Every neighborhood $U$ of the limit contains $a_{2n}$ for every $2n>N$($N$ is an integer). In the same way we find $M$ such that $U$ contains $a_{2n+1}$ for every $2n+1>M$. Taking the maximum of the two integers $M$, $N$, can you prove it now? – William Sun Dec 14 '18 at 00:56
  • You can't come up with a counterexample because is true. A proof can go in the lines of "It is immediate from definition." I am just kidding, it is really easy to prove, see @WilliamSun comment above mine. – William M. Dec 14 '18 at 01:02

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Suppose

$$\lim_{n\to\infty}a_{2n}=\lim_{n\to\infty}a_{2n+1}=A$$

And let $\;\epsilon >0\;$ be arbitrary, then there exist $\;N_1,\,N_2\in\Bbb N\;$ s.t.

$$\begin{cases}|a_{2n}-A|<\epsilon\,,\,\,\forall n>N_1\\{}\\|a_{2n+1}-A|<\epsilon\,,\,\,\forall\,n>N_2\end{cases}$$

Let now $\;M:=\max\{N_1,\,N_2\}\;$ , then for any $\;n>M\;$ , since $\;n\;$ is either even or odd....finish now with one line the proof.

DonAntonio
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