Let $\{b_n\}_{n \in \mathbb{N}}$ be a sequence of complex numbers with only one accumulation point $b$. Let $b_n \neq 0$ for all $n \in \mathbb{N}$. Show that if $0$ is not an accumulation point for $\{1/b_n\}_{n \in \mathbb{N}}$, then $\{b_n\}_{n \in \mathbb{N}}$ converges to $b$.
I know that a bounded complex sequence with only one accumulation point is convergent, and have tried using this result to prove the statement, but I haven't succeeded yet. It keeps me up at night.
since $0$ is not an accumulation point, there exists $c>0$, an integer $N$ such that for every $n>N, |{1\over b_n}|>c$, this implies that $|b_n|<{1\over c}$ and $(b_n)$ is bounded.
Suppose that it does not converge towards $b$, there exists $d>0$, such that for every $p$, there exists $n_p>n$ such that $|b_{n_p}-b|>d$, since $(b_{n_p})$ is bounded, it has an accumulation point $b'=b$, $|b-b'|=|b-b_{n_p}+b_{n_p}-b'|\geq |b-b_{n_p}|-|b'-b_{n_p}|\geq d-|b'-b_{n_p}|$ we can find $n_{p'}$ such that $|b'-b_{n_{p'}}|<{d\over 2}$, we deduce that $0=|b-b'|\geq d-{d\over 2}$ contradiction.