0
$\begingroup$

Prove: If a sequence converges, then every subsequence converges to the same limit.

Instead of saying that $n_{k}\geq k> N\implies |a_{n_{k}}-L|<\epsilon$

Can i say that

let $i$ be the smallest positive integer such that $n_{i}\geq N$,

so $n_{k}> n_{i}\geq N\implies |a_{n_{k}}-L|<\epsilon$

$\endgroup$
4
  • $\begingroup$ Didn't you just change the $k$ with an $n_i$? They basically mean the same thing (index of element/element, that is sufficiently close to $L$) $\endgroup$
    – Laray
    Jan 10, 2017 at 9:23
  • $\begingroup$ @Laray They don't always mean the same thing. It depends on the sequence and subsequence. $\endgroup$
    – Little Rookie
    Jan 10, 2017 at 9:39
  • $\begingroup$ When you take the limit as $n_k\to\infty$, It's sufficient to find one and not necessary be smallest. $\endgroup$
    – Nosrati
    Jan 10, 2017 at 9:54
  • $\begingroup$ @MyGlasses yes, it does not have to be the smallest, but that's not the point here. $\endgroup$
    – Little Rookie
    Jan 10, 2017 at 10:00

0

Reset to default

You must log in to answer this question.

Browse other questions tagged .