Question #204299

Prove that the sequence {an/n} is convergent where { an} is a bounded sequence


1
Expert's answer
2022-01-10T18:24:05-0500

As ana_n is bounded, we know that there exists a positive real number MM such that anM|a_n|\leq M for all nNn\in \mathbb{N}. We will prove that the sequence (ann)nN(\frac{a_n}{n})_{n\in\mathbb{N}} converges to zero, let us fix ε>0\varepsilon>0. We know that there exists NNN\in\mathbb{N} such that for all nNn\geq N, Mn<ε\frac{M}{n}< \varepsilon (it is enough to take N=smallest integer greater than M/εN=\text{smallest integer greater than } M/\varepsilon). We have then an estimate for any nNn\geq N :

annMn<ε|\frac{a_n}{n}|\leq |\frac{M}{n}|<\varepsilon

Therefore, (an/n)nN0(a_n/n)_{n\in\mathbb{N}}\to 0.


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