convergence divergence - If a sequence $(b_n)$ converges to $b$, $b ne 0$, prove that $(frac{1}{b_n})$ converges to $frac{1}{b}$

$(b_n) \rightarrow b$ if $\forall \epsilon > 0$ $\exists N \in \mathbb{N}$ such that if $n > N$, then $|b_n - b| < \epsilon$.

$|b_n - b| < \epsilon$

$|b_n b| \left|\dfrac{1}{b_n} - \dfrac{1}{b}\right| < \epsilon$

$\left|\dfrac{1}{b_n} - \dfrac{1}{b}\right| < \dfrac{\epsilon}{|b_n| |b|}$

If $(b_n)$ is convergent, then it is also bounded : $\exists M > 0 : M \ge |b_n| \forall n \in \mathbb{N}$, this leads to

$\left|\dfrac{1}{b_n} - \dfrac{1}{b}\right| < \dfrac{\epsilon}{M |b|}$

Convergence proofs that I have read so far try to manipulate the expression above into

$\left|\dfrac{1}{b_n} - \dfrac{1}{b}\right| < \epsilon$

Is this necessary? $\dfrac{\epsilon}{M |b|}$ is just another "$\epsilon$" for the $\left|\dfrac{1}{b_n} - \dfrac{1}{b}\right|$ sequence.

I could say: choose such $N$ so that $|b_n - b| < \epsilon M |b|$, which results in

$\left|\dfrac{1}{b_n} - \dfrac{1}{b}\right| <\dfrac{\epsilon M |b|}{M |b|}$

$\left|\dfrac{1}{b_n} - \dfrac{1}{b}\right| < \epsilon$

However, nothing in the convergence theorem states that $\epsilon$ of two such sequences need to be the same. My question: Is it valid to state

$\dfrac{\epsilon}{M |b|} = \epsilon'$ and use the convergence definition?

If $\forall \epsilon' > 0$ $\exists N \in \mathbb{N}$ such that if $n > N$, then $\left|\dfrac{1}{b_n} - \dfrac{1}{b}\right| < \epsilon'$, then $\left(\dfrac{1}{b_n} - \dfrac{1}{b}\right)$ converges.

Since

$\left|\dfrac{1}{b_n} - \dfrac{1}{b}\right| < \dfrac{\epsilon}{M |b|} = \epsilon'$ is equivalent to

$|b_n - b| < \epsilon$

this proves the statement.

Answer

I stopped reading when I saw "this leads to ..."; the inequality there is wrong.

Indeed, to prove the statement, let $b_{n},b \neq 0$ for all $n \geq 1$. If $n \geq 1$, then
$$

\bigg| \frac{1}{b_{n}} - \frac{1}{b} \bigg| = \frac{|b_{n}-b|}{|b_{n}||b|};
$$
there is some $N_{1} \geq 1$ such that $||b_{n}| - |b|| \leq |b_{n}-b| < |b|/2$ for all $n \geq N_{1}$ by convergence of $(b_{n})$, implying that $|b|/2 < |b_{n}|$ for all $n \geq N_{1}$, implying that
$$
\frac{|b_{n} -b|}{|b_{n}||b|} \leq \frac{2|b_{n}-b|}{b^{2}}
$$
for all $n \geq N_{1}$; moreover, given any $\varepsilon > 0$, by convergence of $(b_{n})$ again there is further some $N_{2} \geq 1$ such that $n \geq N_{2}$ implies $|b_{n}-b| < b^{2}\varepsilon/2$, showing that
$2|b_{n}-b|/b^{2} < \varepsilon$ for all $n \geq N_{2}$. Putting all the previous things together, we conclude that, for every $\varepsilon > 0$, we having $n \geq \max \{N_{1}, N_{2} \}$ implies
$$
\bigg| \frac{1}{b_{n}} - \frac{1}{b} \bigg| < \varepsilon.

$$