
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.

1) Let $\displaystyle (s_n)$ be a sequence such that $\displaystyle (\vert s_n \vert )$ converges. Clearly $\displaystyle (s_n)$ must have an infinite number of positive (or negative) terms. Take $\displaystyle (s_{n_k})$ such that $\displaystyle 0 \leq s_{n_k}$ for all $\displaystyle n_k$ and $\displaystyle n_k < n_{k+1}$ then $\displaystyle (s_{n_k})$ is a subsequence of $\displaystyle (s_n)$, but $\displaystyle \vert s_n \vert$ converges.
2)Use the fact that the odd and even numbers partition $\displaystyle \mathbb{N}$

more explicit for 2)
I think for 2) more explicitly it is best to say:
Given $\displaystyle \epsilon >0$, since the even and odd sequences converge to L, there should be an $\displaystyle N_{odd}$ and an $\displaystyle N_{even}$ such that for all odd n bigger than $\displaystyle N_{odd}$ you have $\displaystyle x_nL<\epsilon$ and similarly for all even n bigger than $\displaystyle N_{even}$ you have $\displaystyle x_nL<\epsilon$.
Then let $\displaystyle N:=max\{N_{even},N_{odd}\}$ and you have for all $\displaystyle n>N$ that $\displaystyle x_nL<\epsilon$ so the sequence converges to L as desired.

For the first question I don't think it's possible. For if $\displaystyle s_n$ were bounded, it would have a convergent subsequence by BolzanoWeierstrass. Thus the sequence must be unbounded. But this too isn't possible because if $\displaystyle s_n$ converges we have that by $\displaystyle s_n\le s_n\le s_n$ (nothing is lost by assuming all terms positive really) that $\displaystyle s_n$ would converge. But $\displaystyle s_n$ is unbounded and thus diverges, and so must $\displaystyle s_n$.