# Thread: Sequence Question (not too hard)

1. ## Sequence Question (not too hard)

Hello fellow mathematicians I hope that the summer hasn't rusted your brains here is a question:

Let $\{ x_n \}$ be a sequence and $L \in \mathbb{R}$.
Suppose that $\{ x_{2n} \}$ and $\{ x_{2n-1} \}$ both converge to $L$.
Prove that $\{ x_n \}$ also converges to L.

Is this the correct way of doing this? I'm quite rusty at the moment and any help would be appreciated.

We know the following to be true:

• $\forall \epsilon > 0 \exists N_1 > 0$such that $n > N_1 \implies |x_{2n} - L | < \epsilon$
• $\forall \epsilon > 0 \exists N_2 > 0$such that $n > N_2 \implies |x_{2n-1} - L | < \epsilon$
Let $u = 2n$ and $v = 2n - 1$. $u,v \in \mathbb{N}$ since $2n > 0, 2n - 1 > 0 \forall n \in \mathbb{N}$
So we get:

• $\forall \epsilon > 0 \exists N_1 > 0$such that $\frac{u}{2} > N_1 \implies |x_{u} - L | < \epsilon$
• $\forall \epsilon > 0 \exists N_2 > 0$such that $\frac{v+1}{2} > N_2 \implies |x_{v} - L | < \epsilon$
which after replacing variables and moving the inequality, is equivalent to :

• $\forall \epsilon > 0 \exists N_1 > 0$such that $n > 2N_1 \implies |x_{n} - L | < \epsilon$
• $\forall \epsilon > 0 \exists N_2 > 0$such that $n > 2N_2 - 1 \implies |x_{n} - L | < \epsilon$
Let $\epsilon > 0$. Choose $N = \max \{2N_1,2N_2 - 1\}$. Let $n \in \mathbb{N}$. $n > N \implies |x_n - L| < \epsilon$ .
Is this proof any good? I don't really know I kind of just scribbled it down but if there is anythign wrong I would like to know thanks for the help

2. Yes, that looks completely correct.