# Convergence of Odd and Even Subsequences to Same Limit

## Theorem

Let $\sequence {s_n}$ be a real sequence.

Let the subsequences $\sequence {s_{2 n} }$ and $\sequence {s_{2 n + 1} }$ both converge to the same limit $l$.

Then $\sequence {s_n}$ also converges to the same limit $l$.

## Proof

Suppose that $\sequence {s_n}$ converge to a limit.

Then from Limit of Subsequence equals Limit of Real Sequence, $\sequence {s_{2 n} }$ and $\sequence {s_{2 n + 1} }$ both converge to the same limit.

They do so converge, and that limit is $l$.

So, if $\sequence {s_n}$ converges, it converges to the limit $l$.

Aiming for a contradiction, suppose $\sequence {s_n}$ does not actually converge to a limit.

Then it is not the case that:

- $\forall \epsilon \in \R_{>0}: \exists N \in \R_{>0}: k > N \implies \size {s_k - l} < \epsilon$

That is:

- $\exists \epsilon \in \R_{>0}: \forall N \in \R_{>0}: \exists k > N: \size {s_k - l} > \epsilon$

Either:

- $k = 2 n$ for some $n \in \N$

or:

- $k = 2 n + 1$ for some $n \in \N$

In the first case that means $\sequence {s_{2 n} }$ does not converge to $l$.

In the second case that means $\sequence {s_{2 n + 1} }$ does not converge to $l$.

Both cases contradict the supposition that $\sequence {s_{2 n} }$ and $\sequence {s_{2 n + 1} }$ both converge to $l$.

Hence by Proof by Contradiction it must be the case that $\sequence {s_n}$ converges to a limit.

That limit has been shown to be $l$.

$\blacksquare$

## Sources

- 1975: W.A. Sutherland:
*Introduction to Metric and Topological Spaces*... (previous) ... (next): $1$: Review of some real analysis: Exercise $1.5: 12$