# Even and odd subsequence

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.)

• For the last question, no, consider $(-1)^n$. Sep 15 '20 at 20:25
• @player3236, Thank you. I edited my question. Sep 15 '20 at 20:27

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 $$L$$.