# Convergent sequences

• Jul 16th 2009, 09:47 AM
poorna
Convergent sequences
1) Can you find a sequnce of real nos. {sn} which has no convergent subsequence and yet for which {|sn|} converges.

I am not able to think of such a sequence.:(

2) If a subsequence of {sn} of terms with even subscripts converges to L, as well as the subsequence with odd subscripts (converges to L i.e.) then, prove that the given sequence converges to L.

Here I intuitively know the theorem statement is sort of obvious, but how do I go about the proof?

Any help will be appreciated.
• Jul 16th 2009, 10:09 AM
Jose27
1) Let $(s_n)$ be a sequence such that $(\vert s_n \vert )$ converges. Clearly $(s_n)$ must have an infinite number of positive (or negative) terms. Take $(s_{n_k})$ such that $0 \leq s_{n_k}$ for all $n_k$ and $n_k < n_{k+1}$ then $(s_{n_k})$ is a subsequence of $(s_n)$, but $\vert s_n \vert$ converges.

2)Use the fact that the odd and even numbers partition $\mathbb{N}$
• Jul 16th 2009, 10:34 AM
Gamma
more explicit for 2)
I think for 2) more explicitly it is best to say:

Given $\epsilon >0$, since the even and odd sequences converge to L, there should be an $N_{odd}$ and an $N_{even}$ such that for all odd n bigger than $N_{odd}$ you have $|x_n-L|<\epsilon$ and similarly for all even n bigger than $N_{even}$ you have $|x_n-L|<\epsilon$.

Then let $N:=max\{N_{even},N_{odd}\}$ and you have for all $n>N$ that $|x_n-L|<\epsilon$ so the sequence converges to L as desired.
• Jul 16th 2009, 03:54 PM
putnam120
For the first question I don't think it's possible. For if $s_n$ were bounded, it would have a convergent subsequence by Bolzano-Weierstrass. Thus the sequence must be unbounded. But this too isn't possible because if $|s_n|$ converges we have that by $-|s_n|\le s_n\le |s_n|$ (nothing is lost by assuming all terms positive really) that $s_n$ would converge. But $s_n$ is unbounded and thus diverges, and so must $|s_n|$.