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Suppose $\{a_n\}_{n=1}^{\infty}$ is a bounded sequence of real numbers. Suppose $\lim_{n\to\infty} (a_{2n}+2a_n)=0$. Prove that $\lim_{n\to\infty} a_n=0$.

What I learned from class:

Since $\{a_n\}_{n=1}^{\infty}$ is bounded, $\limsup a_n=\alpha\in \mathbb{R}$, and $\alpha$ has the property of being the largest possible subsequential limit of the $a_n$'s. Furthermore, there exists a subsequence $a_{n_k}\to \alpha$.

We want to show $\alpha=0$, so suppose $\alpha>0$ Now,

$\lim a_{2n_k}=\lim (a_{2n_k}+2a_{n_k})-2\lim a_{n_k}=0-2\alpha=-2\alpha$

$\lim a_{4n_k}=\lim (a_{4n_k}+2a_{2n_k})-2\lim a_{2n_k}=0-2(-2\alpha)=4\alpha$

So we found a subsequence with a limit greater than $\alpha$, so this contradicts $\alpha>0$. We can do a similar argument and conclude $\alpha<0$ is also impossible, so we must have $\alpha=0$.

I have gotten this far, but still cannot conclude that $\lim a_n=0$. How do I even know that the $\{a_n\}$ converge?

Jungleshrimp
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1 Answers1

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Your argument shows that $\limsup_na_n=0$. Now use a very similar argument to show that $\liminf_na_n=0$. Then $\lim_na_n$ exists and is zero.

Martin Argerami
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