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Whe had the following theorem in class:

If $(a_{2n})_{n\in\mathbb N}$ and $(a_{2n+1})_{n\in\mathbb N}$ are convergent sequences with the same limit $a$, then the sequence $(a_{n})_{n\in\mathbb N}$ converges.


  1. $(a_{2n})_{n\in\mathbb N}$ converges, so $\forall\varepsilon>0\exists N_1\in\mathbb N:\forall n\geq N_1: |a_{2n}-a|<\varepsilon$

    It follows $|a_m-a|<\varepsilon$ for all even $m\geq2N_1$.

  2. $(a_{2n+1})_{n\in\mathbb N}$ converges, so $\forall\varepsilon>0\exists N_2\in\mathbb N:\forall n\geq N_2: |a_{2n+1}-a|<\varepsilon$

    It follows $|a_m-a|<\varepsilon$ for all odd $m\geq2N_2+1$.

Thus for all $m\in\mathbb N$ with $m\geq M:=\max\{2N_1,2N_2+1\}$ is $|a_m-a|<\varepsilon$ $\;\;\;\;\;\;\;\;\;\;\;$ q.e.d.

Now, after trying to proove it myself again, I've just wondered myself about one thing:

Why can't you choose $M=\max\{N_1,N_2\}$ and leave the second line in each number out? Why do you have to choose $2N_1$ and $2N_2+1$ ?

Thanks for helping!

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I am sure that it have been already proved here. The idea is the following: consider a subsequence that converges to $\limsup a_n$ - it either have infinitely many odd terms, or infinitely many even terms, thus it converges to $a$. The same holds for $\liminf a_n$, thus the whole sequence converges. – S.D. Nov 30 '12 at 16:27
up vote 1 down vote accepted

You need the factor of $2$ because you drop it from the index. Say, if $N_1=1000$, then this says that $a_{2000}, a_{2002},a_{2004},\ldots$ are close to $a$. Therefore your $M$ needs to be at least $2000$ because $a_{1000}$ may be too far away from $a$.

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so you can't choose the maximum of $N_1$ and $N_2$ - good explanation! – user32778 Nov 30 '12 at 16:38

Since you index the elements of $a_{2n}$ by $n$, saying that $\forall n\geq N_1: |a_{2n}-a|<\varepsilon$ is equivalent to saying that for all $m>2N_1$ even $|a_m-a|<\varepsilon$. It is similar for the second subsequence.

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