Limits of Subsequences and Convergence of Sequences
Theorem
Suppose a sequence $\left\{ a_{n} \right\}$ is given. For the two subsequences $\left\{ a_{2n} \right\}$ and $\left\{ a_{2n+1} \right\}$, if $\lim\limits_{n \to \infty} a_{2n} = L$ and $\lim\limits_{n \to \infty} a_{2n+1} = L$, then $\lim\limits_{n \to \infty} a_{n} = L$.
Proof
By the definition of the limit of a sequence, for every $\epsilon \gt 0$, there exists $N_{1}(\epsilon) \in \mathbb{N}$ satisfying the following.
$$ 2n \ge N_{1} \implies |a_{2n} - L| < \epsilon $$
Likewise, for every $\epsilon \gt 0$, there exists $N_{2}(\epsilon) \in \mathbb{N}$ satisfying the following.
$$ 2n+1 \ge N_{2} \implies |a_{2n+1} - L| < \epsilon $$
Now let $N(\epsilon) = \max(N_{1}(\epsilon), N_{2}(\epsilon))$. Then for every $\epsilon$ the following holds.
$$ n \ge N \implies |a_{n} - L| < \epsilon $$
Therefore, by the definition of the limit of a sequence, we obtain the following.
$$ \lim\limits_{n \to \infty} a_{n} = L $$
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