• Nov 13th 2007, 11:13 AM
There is a sequence, <Xn>, such that the subsequence of even subscripts and odd subscripts both converge to x.

How would I show, using this, that the whole sequence <Xn> converges to x?
• Nov 13th 2007, 11:35 AM
Jhevon
Quote:

There is a sequence, <Xn>, such that the subsequence of even subscripts and odd subscripts both converge to x.

How would I show, using this, that the whole sequence <Xn> converges to x?

well, the sequence $\displaystyle \{ x_n \}$ is consists of terms of even and odd subscripts only. so saying the even and odd subscripts converge, is saying that every term of the sequence converges. but, for the sake of being more formal, we could do this. i'll start you off

Let $\displaystyle \{ x_n \}$ be a sequence. we denote the terms with even subscripts as $\displaystyle x_{2k}$ and those with odd subscripts as $\displaystyle x_{2k + 1}$. if we let $\displaystyle m = 2k$, then the even terms are $\displaystyle x_m$ and the odd terms are $\displaystyle x_{m + 1}$

Now, since $\displaystyle x_m \to x$ we have that for all $\displaystyle \epsilon > 0$ there exists an $\displaystyle N_1 \in \mathbb{N}$ such that $\displaystyle m \ge N_1$ implies $\displaystyle |x_m - x|< \frac {\epsilon}2$. Similarly, for some $\displaystyle N_2 \in \mathbb{N}$, we have $\displaystyle m + 1 \ge N_2$ implies $\displaystyle |x_{m + 1} - x|< \frac {\epsilon}2$

Now take such an $\displaystyle \epsilon$ and choose $\displaystyle N = \mbox{max} \{ N_1, N_2 \}$. Then $\displaystyle m, m + 1 \ge N$ implies that:

$\displaystyle |x_m - x_{m + 1}| = |x_m - x + x - x_{m + 1}| \le |x_m - x| + |x_{m + 1} - x| < \frac {\epsilon}2 + \frac {\epsilon}2 = \epsilon$

That is, we have $\displaystyle |x_m - x_{m + 1}| < \epsilon$

Now, how would you continue. (we want to show that the sequence x_n converges. if that happens, then all its subsequences converge to the same limit as the sequence, which in this case is x. Hint: try to show x_n is a Cauchy sequence)