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Let $a_n$ be a sequence. Assume that $b_n$ and $c_n$ are the odd and even subsequences of $a_n$, respectively. How we can define odd and even subsequences of $b_n$ and $c_n$? I think if $n\equiv 1\pmod 4$ and $n\equiv 3\pmod 4$, then we have odd and even subsequences of $b_n$, respectively. By similar argument, If $n\equiv 0\pmod 4$ and $n\equiv 2\pmod 4$, then we have even and odd subsequences of $c_n$, respectively.

Questions:

  1. Is the above correct?
  2. If i prove the odd and even subsequences of $b_n$ are convergent to the same limit, then $b_n$ is convergent? (Like when the odd and even subsequences of a sequence are convergent to the same limit then the sequence is convergent.)

Thank you for your help.

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    $\begingroup$ For the last question, no, consider $(-1)^n$. $\endgroup$
    – player3236
    Sep 15 '20 at 20:25
  • $\begingroup$ @player3236, Thank you. I edited my question. $\endgroup$
    – N math
    Sep 15 '20 at 20:27
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  1. Consider the sequence $a_n = n$.

Then $(b_n)=1,3,5,7,9,\dots$ and $(c_n) = 2,4,6,8,10,\dots$.

The odd subsequence of $c_n$ is therefore actually $2, 6, 10,\dots$ which is $n\equiv 2\pmod 4$.


  1. Yes. Any two complementary subsequences converging to the same limit implies the convergence to the same limit of the original sequence:

Let $\epsilon > 0$. Suppose both $b_{2n+1}$ and $b_{2n}$ converges to $L$. Then:

$$\exists M\in\mathbb N: \forall 2n+1>M:|b_{2n+1}-L|<\epsilon$$ $$\exists N\in\mathbb N: \forall 2n>N:|b_{2n}-L|<\epsilon$$

Hence for all $n>\max\{M,N\}$, $|b_n - L|<\epsilon$, which proves that $b_n$ converges to $L$.

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