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Let {a_n} be a sequence such that lim a_2k = L = lim k→∞ a_(2k+1) Prove that lim n→∞ a_n = L.

  • I know that I'll have two limits set up, |a_2k - L| < ε and |a_(2k+1)-L| < ε and then I'll have two inequalities, k ≥ N_1 that comes from n = 2k and k ≥ N_2 that comes from n = 2k+1. So in total I'll have two cases. I just don't know how to fill in my proof with the language it requires to make sense. Can someone please help me?
Arctic Char
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Sam
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1 Answers1

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sorry I guess this should be a comment and not an answer.

As you said, you have a sequence $(a_n)_{n \in \mathbb{N}}$ satisfying that for all $\epsilon > 0$ exists $n_0 \in \mathbb{N}$ such that for all $n > n_0$ we get $\vert a_n - L \vert < \epsilon$.

One way to understand this is by saying:
"For the sequence to converge towards $L$, infinitely many elements of the sequence need to be contained within an arbitraryly narrow tube of radius $\epsilon$, and only finitely many are allowed outside."

Furthermore this means that the $n_0$ in the definition can be understood as the last element of the sequence which is not contained within the $\epsilon$-tube. (Try to picture it.)

In your case you have split your sequence in two, and for both parts you get that for all $\epsilon$ there exists a $n_1$ for the even elements, and an $n_2$ for the odd ones. Using the intuition above, there is a pretty easy and obvious way to get from these two the $n_0$ corresponding to $\epsilon$ needed to prove convergence of the sequence $(a_n)$.

Tobi S
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