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If a sequence $\{x_n\}$ has the property that $$\lim_{n\rightarrow\infty} x_{2n} = \lim_{n\rightarrow\infty}x_{2n+1} = L$$ Show that the sequence $\{x_n\}$ convergences to $L$.

I tried a proof by contradiction against the given information, then I considered saying that if all the subsequences of the already given convergent subsequences converge to $L$, then $\{x_n\}$ must converge to $L$.

What is the proper way to prove this?

Aspiring Mathematician
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2 Answers2

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Hints: for all $\;\epsilon >0\;$

$$\begin{align*}\exists\,M_1\in\Bbb N\;\;\;s.t.\;\;n>M_1\implies& |x_{2n}-L|<\epsilon\\ \exists\,M_2\in\Bbb N\;\;\;s.t.\;\;n>M_2\implies& |x_{2n+1}-L|<\epsilon\end{align*}$$

Well, now just take a peek at $\;M:=\max(M_1,M_2)\;\;$ ...

DonAntonio
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The fact that $\lim_{n\rightarrow\infty} x_{2n} = \lim_{n\rightarrow\infty}x_{2n+1} = L$ implies

$(\forall \epsilon > 0)(\exists n_0 \in \mathbb N)(\forall n>n_0)(|x_{2n}-L|<\epsilon$ and $(\forall \epsilon > 0)(\exists n_1 \in \mathbb N)(\forall n>n_1)(|x_{2n+1}-L|<\epsilon$.

In other words, from $n_0$ onwards, distance of even terms from $L$ is smaller than any $\epsilon$ and from $n_1$ onwards distance of odd terms from $L$ is smaller than any $\epsilon$.

Define $n_2:=\max(n_0, n_1)$. For all $n$ bigger than $n_2$, distance of both even and odd terms is smaller than any $\epsilon$. In other words,

$(\forall \epsilon > 0)(\forall n>n_2)(|x_{n}-L|<\epsilon)$

which is equivalent to $\lim_{n\rightarrow\infty} x_{n} = L.$

mirgee
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