Verification of proof of the Sequence of Arithmetic Theorem Suppose $\left\{b_{n}\right\}$ is a sequence of real numbers which converges to $M$, so that $b_{n} \neq 0$ for each $n$, and $M \neq 0$. Prove that the sequence $\{ \frac{1}{b_n} \}$ converges to $\frac 1M$ by using the following steps:
(a) Prove that there is a $\delta > 0$ so that $| b_{n} | \geq \delta$ for all $n$.  (Hint: consider $\epsilon = \frac{|M|}{2}$.)
(b). Prove that $\frac{1}{| b_n M |}  < \frac{2}{\delta | M |}$.
(c). Prove that $\{ \frac{1}{b_n} \}$ converges to $\frac{1}{M}$. (Hint: use the fact that $| \frac{1}{b_n} - \frac{1}{M} | = 1/\left\vert b_{n}M\right\vert\left\vert b_{n} - M\right\vert$.
For part (a): First we prove that ${|b_n|}$ converges to $|M|$. Second we explicitly show that a $\delta$ > 0 with $| b_{n} | \geq \delta$ for all $n$.
Let $\epsilon > 0$. Since the sequence $\left\{b_{n}\right\}$ converges to $M$, there is an $N$ is an element of the set of real numbers, so that if $n$ belongs to the set of natural numbers $n$> $N$ then $|b_n - M|$ < $\epsilon$. Suppose that $n$ is in the set of natural numbers and $n$ > $N$. Then by the triangle inequality, we have $||b_n| - |M||\leq \ |b_n - M|$ < $\epsilon$. Thus, the sequence ${|b_n|}$ converges to $|M|$. 
For the second part, we let $\delta = \frac{|M|}{2}$. Applying the definition of convergence with $\epsilon = \frac{|M|}{2}$ gives us an $N$ which belongs to a set of reals, so that if $n$ belongs to the set of natural numbers and $n$ > $N$ then $||bn| - |M||< |M|/2$. Then, if $n \gt N$ we have $-|M|/2 \lt bn \lt |M|/2$. Adding $|M|$ gives $|M|/2 \lt b_n \lt M + |M|/2$. In particular, if $n \gt N$, then $|b_n| \gt |M|/2 = \delta$ . 
For part (b): $\frac{1}{| b_n M |}$ = $\frac{1}{| b_n| |M| }$ > $\frac{1}{\delta |M| }$ < $\frac{2}{\delta |M|}$ (since $\frac{1}{|M|}$ < $\frac{2}{|M|}$).
For part (c): Since the sequence $\left\{b_{n}\right\}$ converges to $M$  ≠ 0, by part a) there is an $N_0$ in $R$ (the set of real numbers) and $\delta$ > 0 so that if $n$ belongs to the set of natural numbers and $n$ > $N$ then $| b_{n} | \geq \delta$. Then if $n$ > $N_0$, $\frac{1}{|b_{n}|}$ < $\frac{1}{\delta}$. Let $\epsilon$ > 0. Since sequence $\left\{b_{n}\right\}$ converges to $M$, there is an $|N_b|$ in $R$ so that if n belongs to the set of natural numbers and $n$ > $N_b$ then $|b_n - M|$ < $\epsilon$ $\delta$ $|M|$. let $N$ = max($N_0$, $N_b$). Suppose that $n$ belongs to the set of naturals and $n$ > $N$. Then $| \frac{1}{b_n} - \frac{1}{M} | = 1/\left\vert b_{n}M\right\vert\left\vert b_{n} - M\right\vert$ = $\frac{|b_n - M|}{|b_n| |M|}$ < $\frac{\epsilon\delta |M|}{|b_n| |M|}$ = $\epsilon$. Hence, $\{ \frac{1}{b_n} \}$ converges to $\frac{1}{M}$.
So what you guys think about my proof?
 A: I don't know if it is by accident or by error, but you seem to have mixed at least $b_n$ and $b_n- M$ a few times. How about this approach for the first part:
$|M|\le |b_n|+|M-b_n|$ by the triangle inequality. As you noted, $|M-b_n|<\frac12|M|$ for almost all $n$, hence $|b_n|\ge |M|-|M-b_n|>(1-\frac12)|M|$ for almost all $n$.
A: This is only for part $(a)$.  First prove that if a sequence $b_n \rightarrow M$, then any subsequence also converges to $M$.  This follows since $\forall n \gt N, |b_n - M| \lt \epsilon$ which includes those $b_n$ in the subsequence.  If you want to formalize it more, a subsequence is a sequence $(b_{k_n})$ such that $k_i \lt k_j$ for $i \lt j$ and you also have that $k_n \geq n$.
Argue by contradiction.  If there exist no such $\delta$ intuitively it seems that $b_n$ or subsequence might converge to zero, but since $M \neq 0$, there's a contradiction if a subsequence does.  So we want to construct such a subsequence.  If no such $\delta$ exists, then for $\delta = 1/m$ there exists infinitely many $b_n$ less than less than $\delta$.  Choose one and call it $b_{k_m}$.  Then $(b_{k_m})_{m\geq 1}$ is a subsequence converging to zero since for all $\epsilon \gt 0$...  
Something like that.  Sorry I didn't answer w.r.t. the hint for part $(a)$, and I don't know if the above method is valid.  But you should look into it and let me know how it goes.
