Question

Could somebody help me to understand a proof concerning limits of sequences.

Answer 1

Proof in the attachment

Answer 2

I don't understand why \[|b_k-L|<\epsilon\] How to arrive to this conclusion?

Answer 3

by definition, \[\lim f(x)=L\] if \[\left| f(x) - L \right|< \varepsilon\] this is the construct of the proof (of a limit).

Answer 4

here's the problem if i interpreted it right. Prove that a subsequence of a convergent sequence is convergent, and that both sequence converges to the same value.

Answer 5

assume that a sequence \(\{a_n\}\) converges to some number L. In symbols,\[\lim_{n \rightarrow +\infty}a_n=L\] or equivalently (by definition of a limit), \[\forall \epsilon>0, \exists N>0 \backepsilon n>N \text{ whenever } |a_n-L|<\epsilon \]

Answer 6

There is an \(N\) and an \(\epsilon\) that guarantees \(L\).

Answer 7

To prove that the subsequence, say \(\{b_n\}\), of \(\{a_n\}\) converges to the same limit, you must prove that \[\lim_{n \rightarrow \infty}b_n=L\]

Answer 8

Yeah, I understand that. I don't understand how does \[n_k \ge n_K \ge N \] proove that \[|b_k - L| < \epsilon \]

Answer 9

the criterion \[n>N \text{ whenever } \huge | a_n - L|< \epsilon\]

Answer 10

choose \(n_K\) for \(a_n\)

Answer 11

the trick in the proof is to "beat" the \(\displaystyle n_K\) by being greater than \(n_K\), that is, \(n_k>n_K\), so that you can use the part \(n_k>N\) of the assumption that implies \( |a_{n_K}-L|<\epsilon\)