# Prove, if a sequence converges, its subsequences converge and have the same limit?

• May 3rd 2009, 09:12 PM
qtpipi
Prove, if a sequence converges, its subsequences converge and have the same limit?
Prove that if <a> converges, then every subsequence of <a> converges and has the same limit as a.
• May 3rd 2009, 09:26 PM
TheEmptySet
Quote:

Originally Posted by qtpipi
Prove that if <a> converges, then every subsequence of <a> converges and has the same limit as a.

let $\displaystyle \epsilon > 0$

suppose that $\displaystyle \{a_n\} \to a$

let $\displaystyle \{a_{n_k}\}$ be any subsequnce

Since the sequence convereges there is an $\displaystyle N \in \mathbb{N}$ such that for all n > N

$\displaystyle |a_n-a|< \epsilon$

but since $\displaystyle \{a_{n_k}\}$ if futher in the sequnce then

$\displaystyle \{a_{n}\}$

so $\displaystyle |\{a_{n_k}\}-a|< \epsilon$
• May 3rd 2009, 10:15 PM
qtpipi
Quote:

Originally Posted by TheEmptySet

but since $\displaystyle \{a_{n_k}\}$ if futher in the sequnce then

$\displaystyle \{a_{n}\}$

so $\displaystyle |\{a_{n_k}\}-a|< \epsilon$

I'm not sure what you're saying here, can you rephrase it? Thanks
• May 4th 2009, 05:53 AM
TheEmptySet
Quote:

Originally Posted by qtpipi
I'm not sure what you're saying here, can you rephrase it? Thanks

The integer $\displaystyle n_k \ge n$ a subsequence has all of the same terms as the regular sequence. It just skips some

ie $\displaystyle a_n=\frac{1}{n}$ and

$\displaystyle a_{n_k}=\frac{1}{2^{n_k}}$

the first gives the sequence

$\displaystyle a_n=\left\{ \frac{1}{1} ,\frac{1}{2} ,\frac{1}{3} ,\frac{1}{4} ,\frac{1}{5} ,... \right\}$

but

$\displaystyle a_{n_k}=\left\{ \frac{1}{1} , \frac{1}{2} ,\frac{1}{4} ,\frac{1}{8} ,\frac{1}{16} ,... \right\}$

the subsequce skips terms from the original so to get the term

$\displaystyle \frac{1}{16}$ $\displaystyle n=16$ but $\displaystyle n_k=4$

So $\displaystyle n \ge n_k$