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Let $(a_{n})_{n=1}^{\infty}$ be a bounded sequence, and suppose that the sequence $(a_{n}+ 2a_{n+1})_{n=1}^{\infty}$ converges to $3L \in R$. How can I prove that $(a_{n})_{n=1}^{\infty}$ converges to $L$ ?

How can I go about proving this? I tried using the definition of a limit for the 2 sequences, and tried iterating the inequalities, but I did not seem to be getting anywhere. Could someone provide a proof for this problem, I am really trying my best to understand a proof for this problem.

Also, is there any way to prove the above without, assuming that $(a_{n})_{n=1}^{\infty}$ is bounded.

Really would appreciate your help on this problem. Thank You!

Viktor Raspberry
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  • @WillR Yeah, I fixed the title. Thanks – Viktor Raspberry Jan 30 '18 at 06:59
  • By the referenced link, the fact that $(a_n+2a_{n+1})$ converges implies $(a_n)$ converges, hence $(a_n)$ is bounded. Thus, boundedness of $(a_n)$ is already implied. – quasi Jan 30 '18 at 07:09
  • With the additional initial assumption that the sequence is bounded, you could make an argument starting with letting $M := \limsup a_n$ and $m := \liminf a_n$. Then, looking at elements where $a_{n+1}$ is near the limsup would give you that $m-L \le -2(M-L)$ and similarly looking at elements near the liminf would give you that $M-L \ge -2(m-L)$, so together you get $m-L = M-L = 0$. That argument wouldn't really work to eliminate the possibility $M=\infty$, $m=-\infty$ though. – Daniel Schepler Jan 26 '21 at 00:27

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