If $a_n$ is a sequence such that the sequence $a_{n+2}$ is convergent, does this mean that $a_n$ must also be convergent?

Suppose $a_n$ is a sequence such that the sequence $a_{2n}$ is convergent. Must $a_n$ also be convergent?

I do not know how to even approach these two questions, but I am guessing that it might have something to do with monotone convergence theorem since that particular theorem was simply "given" to me to help me prove that some series are convergent and that their limits exist.

Any help and/or guidance would be much appreciated.

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  • $\begingroup$ The first is yes, because all but finitely many terms of $a_n$ are in the sequence $a_{n+2}$. The second is no, for example my sequence could be $0101010101010101...$. $\endgroup$ – астон вілла олоф мэллбэрг Oct 20 '16 at 7:21
  • $\begingroup$ Also, this has nothing to do with monotone convergence theorem. $\endgroup$ – Ivan Neretin Oct 20 '16 at 7:22

If $a_n$ is convergent then for any fixed $k$, so it $a_{n+k}$. This is fairly straightforward to verify from the definition.

For the other, take the sequence $a_1=1, a_2 = 0, a_3 = 3, a_4 = 0, a_5 = 5, a_6 = 0, ....$. then $a_{2n} = 0 \to 0$ but clearly the sequence does not converge.

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  • $\begingroup$ Thank you for your explanation of the first part. I reread the definition of sequence and it does indeed make a lot of sense. In regards to the second part, I am stiff confused. I apologize, but would you please be able to clarify it for me? $\endgroup$ – Cherry_Developer Oct 20 '16 at 7:31
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    $\begingroup$ It is an example of a sequence that does not converge but the even terms do. $\endgroup$ – copper.hat Oct 20 '16 at 7:36
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    $\begingroup$ Do you know any convergent sequence? Take its terms and put them in even positions; then $a_{2n}$ would converge. Now what does this tell you about the odd terms? Nothing. They can be made anything. Do you know any obviously divergent sequence? Take its terms and put them in odd positions. $\endgroup$ – Ivan Neretin Oct 20 '16 at 7:36

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