If $a_{n+1}-\frac{a_n}{2} \to 0$ then $a_n \to 0$

Question

Assume that $a_{n+1}-\frac{a_n}{2} \to 0$

Prove that $a_n \to 0$

This is my attempt: Let $\epsilon>0$.

Exists $n_0 \in \Bbb{N}$ such that $\frac{a_n}{2}-\epsilon<a_{n+1}<\frac{a_n}{2}+\epsilon,\forall n\geq n_0$

Let $n>n_0$

Then $\frac{a_{n_0}}{2}-\epsilon<a_{n_0+1}<\frac{a_{n_0}}{2}+\epsilon$

$\frac{a_{n_0}}{4}-\frac{\epsilon}{2}-\epsilon<a_{{n_0}+2}<\frac{a_{n_0}}{4}+\frac{\epsilon}{2}+\epsilon$

$\frac{a_{n_0}}{8}-\frac{\epsilon}{4}-\frac{\epsilon}{2}-\epsilon<a_{{n_0}+3}<\frac{a_{n_0}}{8}+\frac{\epsilon}{4}+\frac{\epsilon}{2}+\epsilon$

Preoceeding this way $n-n_0-$times,we have that $$\frac{a_{n_0}}{2^{n-n_0}}-\sum_{i=1}^{n-n_0}\frac{\epsilon}{2^{i-1}}<a_n<\frac{a_{n_0}}{2^{n-n_0}}+\sum_{i=1}^{n-n_0}\frac{\epsilon}{2^{i-1}}$$

From this we deduce that $|\limsup_na_n|\leq\epsilon$ and $|\liminf_na_n|\leq \epsilon$

Thus $a_n \to 0$

Is my strategy correct or am i missing something?

Also i would like to know if there are other solutions for this problem.

Thank you in advance.

Answer 1

It looks good. There are a few small problems:

1. In your sums, you should have written $\displaystyle\sum_{i=1}^{n-n_0}\frac\varepsilon{2^{i-1}}$ instead of $\displaystyle\sum_{i=1}^{n-n_0}\frac\varepsilon{2^{n-1}}$.
2. You deduced that $|\limsup_na_n|,|\liminf_na_n|\leqslant\varepsilon$. In fact, it should be $2\varepsilon$, not $\varepsilon$.