Let $\{a_{n}\}$ is Cauchy sequence and $\{b_{n}\}$ is a sequence satifying $|a_{n}-b_{n}| < \frac{1}{n}$ for every $n \in \mathbb{N}$ then $\{b_{n}\}$ converges.
I was wondering if someone would verify my work below. I am unsure if every step I took was justifiable, true, or complete. Thank you!
My attempt:
Let $\epsilon > 0$. Now, $(|a_n - b_n|) < \frac{1}{n}$ implies that there exists $N_1 \in \mathbf{N}$ such that $|a_n - b_n| < \epsilon /2$ for all $n \ge N_1$. Now assume $(a_n) \rightarrow a$, since it is Cauchy sequence, hence, converges, which implies that there exists $N_2 \in \mathbf{N}$ such that $|a_n - a| < \epsilon /2$ for all $n \ge N_2$. Set $N = \max\{N_1, N_2\}$, then whenever $n \ge N$ it follows that $|b_n - a| = |b_n - a -a_n+a_n| \le |a_n - b_n| + |a_n-a| < \epsilon/2 + \epsilon/2 = \epsilon$, hence $(b_n) \rightarrow a$.