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How do I prove $\lim_{n\to\infty} a_n = \lim_{n\to\infty} a_{n+k} \text{ and } a_{n+1} $ is convergent, where $k$ is a fixed natural number and $a_n$ is a convergent sequence? I know I need to use the definitions but I'm not sure how to manipulate them.

Edit: I have the idea of the proof but I am not sure how to formally write it.

Chubbles
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  • This is a consequence of the fact that every subsequence of a convergent sequence is convergent and has the same limit. Basically by adding $k$ you are removing the first $k$ terms of the sequence but the infinite tail will remain the same – Luis Vera Mar 01 '16 at 18:50
  • See: http://math.stackexchange.com/questions/213285/prove-if-a-sequence-converges-then-every-subsequence-converges-to-the-same-lim – Luis Vera Mar 01 '16 at 18:50
  • That makes sense. ok – Chubbles Mar 03 '16 at 18:53

2 Answers2

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Hint: If you know that $\vert a_n-L \vert < \varepsilon$ for $n\ge N_\varepsilon$, for which $n$ is it true that $\vert a_{n+k}-L \vert < \varepsilon$?

πr8
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  • n-k? that would make $a_{n+k} = a_n$ Am I allowed to do that? – Chubbles Mar 03 '16 at 18:51
  • I mean for which set of values of $n$ - in this case, you should arrive at $n \ge N_\varepsilon - k$. Does this make sense? – πr8 Mar 04 '16 at 14:47
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Definition: We say that $a_k$ converges to $L$ if and only if for all $\epsilon>0,\exists N$ such that if $n>N$, then $|a_n-L|<\epsilon$.

What you want to do is find an $N_1$ and an $N_2$ such that the definition of convergence is simultaneously satisfied by both sequences for the same $\epsilon$.

Stella Biderman
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  • I get what you're saying, not sure how I would go about this. For example, if $a_n$={1,2,3,4,0,0,0,0,0,0} then $N_1$ = 5 satisfies the definition. If I take k = 3, then $a_{n+k}$ would be at least something like {1,2,3,4,0,0,0,0,0,0,0,0,0} then $N_2 = N_1 - k $ would work? – Chubbles Mar 03 '16 at 19:01
  • That is exactly correct. Apply an identical thing to the general case. – Stella Biderman Mar 03 '16 at 21:37