Given that sequences $(s_n)^\infty_{n=1}$ and $(t_n)^\infty_{n=1}$in the following case, how to prove that $s \le t$?

Question

Given that sequences $(s_n)^\infty_{n=1}$ and $(t_n)^\infty_{n=1}$ converge to limits $s$ and $t$, respectively, and given that $s_n \le t_n$ for every $n \gt 100$, how to prove that $s \le t$?

My procedure: we are given $\forall\epsilon_s>0$ and $\forall\epsilon_t>0, \exists N\in \mathbb{R}$ (respectively) such that $n>N\implies |s_n-s|<\epsilon_s$ and $|t_n-t|<\epsilon_t$. I guess we may be able to use contradiction. Suppose $s \gt t$, then $|s-s_n|+s_n>|t-t_n|+t_n \ \implies \ \epsilon_s+s_n>\epsilon_t +t_n$. Then pick $\epsilon$ so small that they are negligible, so $s_n\gt t_n$ shows a contradiction.

I felt my proof is full of guess and errors. Could someone point them out?

Answer 1

Suppose $ s>t $ Then we choose $\epsilon={s-t}/2$ .The interval $(s-\epsilon,s+\epsilon)$ contains all but finitely many $s_k 's$ and similarly the interval $(t-\epsilon,t+\epsilon)$ contains all but finitely many $t_l 's$ so there exists $N\in \mathbb{N}$ large enough such that $t_N\in(t-\epsilon,t+\epsilon)$ and $s_N\in(s-\epsilon,s+\epsilon)$ this shows $s_N>t_N$ a contradiction.

Answer 2

go for a contradiction seems a good idea ;but where you say suppose s>t then * I don't see how you got the inequality * . But -pick e so small so that e is negligble is a good strategy but is but not precise enough to count in a proof .What e should you use .And another thing, Why not take e_s and e_t to be the same ,Just call it e >0 . Whatever your e is You can find N (the same one for both sequences) . Lets assume tN will do ) But s_n <= t_n .You can see that this means that s-e< t-e or s-t < e e can be any positive number So s-t can not be positive ,otherwise e could be =s-t and we would get e< e ,a clear contradiction Hope that helps .Stuart M.N.

Answer 3

Instead, try proving that if $(a_n)_{n=1}^\infty$ is a sequence which converges to limit $L$ and $a_n\geq 0$ for all $n>100$, then $L\geq 0$.

Proof: Suppose on the contrary that $L<0$. Since $a_n\rightarrow L$, given any $\epsilon>0$ there exists an integer $N$ such that $n>N$ implies $|a_n-L|<\epsilon$. Take, in particular, $\epsilon =\dfrac{-L}{2}$. Then there is an integer $N$ such that $n>N$ implies $$|a_n-L|<\frac{-L}{2}$$ $$\Rightarrow \frac{L}{2}<a_n-L<\frac{-L}{2}$$ $$\Rightarrow L+\frac{L}{2}<a_n<L-\frac{L}{2}=\frac{L}{2}<0$$ that is $a_n<0$ for all $n>N$ which contradicts the fact that $a_n\geq 0$ for $n>100$.

Then apply this result to the sequence $a_n=t_n-s_n$ which converges to $L=t-s$.