Prove that if $(a_n)$ converges and $|a_n - nb_n| < 2$ for all $n \in \mathbb N^+$ then $(b_n)$ converges.

Is the following proof valid?

Since $(a_n)$ converges, $(a_n)$ must be bounded, i.e. $\exists M \in \mathbb R^+$ such that for each $n \in \mathbb N^+$, we have $|a_n| < M$.
Now, by the triangle inequality, $|nb_n|$ = $|nb_n - a_n + a_n| \le |nb_n - a_n| + |a_n| < 2 + M$.
Hence, $|b_n - 0| < \frac{2 + M}{n}$.
Let $\epsilon > 0$ be given, and by the Archimedean Property of $\mathbb R$, we can choose $K \in \mathbb N^+$ such that $K > \frac {2+M}{\epsilon}$.
Then, $n \ge K \implies n > \frac {2 + M}{\epsilon} \implies |b_n - 0| < \epsilon$.
Therefore $(b_n)$ converges, and its limit is $0$.

  • $\begingroup$ Yes. It is correct $\endgroup$ – HK Lee Apr 20 '15 at 4:35

Your proof is correct. In fact, once you have $$|b_n|<\frac{M+2}{n}$$ it is clear that since the numerator is bounded one can make the right hand side $<\epsilon$ for arbitrarily small $\epsilon >0.$


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