Can we deduce that $uLet $u_{n}, v_{n}$ be two convergent real sequences that converge to $u$ and $v$ respectively. Assume that $u_{n}<v_{n}$ for all $n≥2$. My question is: Can we deduce that $$u<v$$
If not, what are the necessary conditions for that property?
 A: Let $u_n = 0, v_n = {1\over n}$. Then $u_n < v_n$ for all $n$, but $\lim_n u_n = 0 = \lim_n v_n$.
A: 
Can we deduce that $u<v$[?]

Of course not, but $u\leqslant v$. Examples where $u=v$ abound, surely you can find some yourself.

If no, what are the necessary conditions for that property.

A necessary and sufficient condition is that $u_n$ and $v_n$ are not too close, in the sense that
$$
\inf_n\ (v_n-u_n)\gt0.
$$
Proving both implications is a good exercise, if you meet some specific difficulties doing so, just yell.
A: If you need $u<v$ then starting from some $n=N$ all $u_n$ should be within, say $\frac{v-u}{3}$-neighborhood of $u$ and similarly for $v_n$, i.e. you need not only $u_n<v_n$, but also something like this: there exists $\epsilon>0$ s.t. starting from some $N$: $n\ge N$ implies $u_n<v_n-\epsilon$.
A: In general you cannot deduce the strict inequality. $u_n < v_n$ implies only $u \leq v$. Imagine two sequences like 
$$ 1-\frac{1}{n}<1+\frac{1}{n}$$
which have the same limit.
One sufficient and necessary condition to deduce that $u<v$ would be to intercalate two values in between: $u_n \leq c<d\leq v_n$ (for $n$ sufficiently large). If you can do that then obviously $u \leq c<d \leq v$ so $u<v$. 
Conversely, if $u<v$ then for $n$ great enough $u_n < u +\frac{v-u}{3}<v-\frac{v-u}{3}<v_n$.
