0
$\begingroup$

Let $(x_n)_{n\in \Bbb N}$ be a sequence in real numbers and a$\in \Bbb R$. If exists m $\in \Bbb N$ so that $lim_{m+n \to \infty} x_n = a$. Prove that $lim_{n \to \infty} x_{n}=a$

It is a problem that, I think it involves the definition of a limit as $\forall \epsilon \gt 0, \exists N \in \Bbb N / |x_n-a|< \epsilon, \forall n\ge N $ is the same as $lim_{n \to \infty} x_n=a$ An answer using this definition would be appreciated, as I have tried to find a moment N so that it becomes true.

$\endgroup$
1
  • $\begingroup$ $\lim_{m+n\to\infty}x_n = \lim_{k\to\infty}x_{k-m}$. $\endgroup$ – amsmath Oct 4 '19 at 18:02
1
$\begingroup$

Given $$ \lim_{m+n \to \infty}x_{n}=a $$ $$ \implies \forall \epsilon >0, \exists N \in \mathbb{N} ~ \text{s.t}~ |x_{n}-a|< \epsilon ~ \text{for} ~ m+n \ge N. $$ $$ \text{Choose}~ M=N-m, ~ \text{then} ~ \forall \epsilon >0, \exists M \in \mathbb{N} ~ \text{s.t} ~ |x_{n}-a|< \epsilon ~ \text{for} ~ n \ge M. $$ $$ \text{Therefore,} ~ \lim_{n \to \infty}x_{n}=a $$

$\endgroup$
0
$\begingroup$

If you want to use the definition it enough to replace $N$ by $N+m$ but the idea is that you only care when $n$ is large, so it doesn't matter if you "drop" the first $m$ terms of the sequence, (the limit won't change).

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.