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Suppose $A_n \to L$ and $B_n \to L$.

I need to show that the sequence $A_1,B_1,A_2,B_2,A_3,B_3,\dots$ converges to $L$. Now, I know that both $A_n$ and $B_n$ are less than or equal to $L$ for all $n$, however, how can I show that if you interleave the items in $A_n$ and $B_n$, it will also converge to $L$? It makes sense, I just don't know how to approach this problem.

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Let $\epsilon>0$. Choose $N_1$ so that for $n> N_1$, $|A_n-L|<\epsilon$. Choose $N_2$ so that for $n> N_2$, $|B_n-L|<\epsilon$. What happens with the interleaved sequence if $n>2\max\{N_1,N_2\}$? –  David Mitra May 10 '12 at 20:24
    
Choose $N = \max(N_A,N_B)$ so that it works for both the sequences. –  user17762 May 10 '12 at 20:24
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You do not know that $A_n,B_n\le L$ for all $n$: the sequences might converge to $L$ from above, or they might bounce above and below $L$ while still converging to it. –  Brian M. Scott May 10 '12 at 20:25
    
@alexthebake: Is the fact that $A_n\leq L$ and $B_n\leq L$ an extra piece of information, or something that you are deducing? You can't deduce it, and it's also not necessary. –  Tara B May 10 '12 at 20:26

2 Answers 2

Informally speaking, you have to show that the terms of the interleaved sequence get closer and closer to $L$ as the index becomes larger and larger. Each of the sequences $(A_n)$ and $(B_N)$ has this property; it seems obvious that the interleaved sequence will have this property.

But that's all rather vague. Formally, you need to show that for any given $\epsilon>0$, there is a positive integer $N$ so that whenever $n>N$, the $n^\text{th}$ term of the interleaved sequence, call it $I_n$, satisfies $|I_n-L|<\epsilon$.

Towards that end, here's a hint:

Let $\epsilon>0$. Choose $N_1$ so that for $n>N_1$, we have $|A_n−L|<\epsilon$. This can be done since $(A_n)$ converges to $L$. Then choose $N_2$ so that for $n>N_2$, we have $|B_n−L|<\epsilon$. This can be done since $(B_n)$ converges to $L$.

Now, what can you say about the $n^{\rm th}$ term of the interleaved sequence if $n>2\max\{N_1,N_2\}$?

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I think that it is rather obvious that if the odd terms in the sequence converge to L and the even terms also converge to L then the sequence has to converge to L. Convergence along the subsequences means for large enough N the difference between An and L will be less than any given epsilon and the same can be said for Bn at some other large value M. So the entire sequence will be less than epsilon away from L anywhere at or beyond T=max(N,M). this is similar to David Mitra's answer.

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