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$(a_n)$ is a bounded sequence such that $$\lim_{n\to\infty}(a_{2^n}-3a_n)=0$$

Prove that $(a_n)$ converges and determine the limit.

I haven't got a clue of what to do here. I've been told I could watch accumulation points of the sequence but honestly I don't understand how I should proceed. Any kind of hint would be really helpful. Thanks in advance!

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    I edit. Is that what you mean? – 2017-01-15
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    @juniven Oh I see, thanks a lot, yes that is what I meant! :) – 2017-01-15

2 Answers 2

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Hint: If $L$ is a limit point then $3L$ is also a limit point. Since sequence is bounded $L=0$. Thus sequence converges to $0$.

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    I see! that makes sense, but (I hope this isn't a too dumb question) how am I to know/prove that the sequence $(a_n)$ is bounded by 0? I can directly conclude that from the limit I've been given or? – 2017-01-15
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    @DomoB: question is not dumb at all! Use recursion. If 3L is limit point then 9L is also limit point and so $3^{n}L$ is also a limit point. If $L\neq0$ then this diverges to infinity. – 2017-01-15
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    Awesome, thanks! So basically my proof says that L has to be 0, for the sequence to converge,because otherwise it would have more than one accumulation point in which case it doesn't converge? Thanks a lot for your help! :) – 2017-01-15
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Hint (for the limit value):

Assume $\lim_{n\to \infty} a_n=s$, plug this into $$\lim_{n\to\infty}(a_{2^n}-3a_n)=0$$ and solve for $s$.