$(b_n) \rightarrow b$ then $(1/b_n) \rightarrow 1/b$ proof.

Question

Let $(b_n)$ be a sequence that converges to $b$. Show that $(1/b_n)$ converges to $1/b$ when $b_n \neq 0$ for all $n$ and $b \neq 0$.

First see that

$$ |\frac{1}{b_n} - \frac{1}{b}| = |\frac{b - b_n}{b_nb}| = \frac{|b_n - b|}{|b_n||b|} $$

Now from convergence of $(b_n)$ it follows that $\forall \epsilon \geq 0$ there is $N \in \mathbb{N}^{+}$ such that $|b_n - b| \leq \epsilon$ when $n \geq N$.

That implies that $\forall \epsilon \geq 0$ when $n \geq N$, it is true that $$ \frac{|b_n - b|}{|b_n||b|} \leq \frac{\epsilon}{|b_n||b|} = \epsilon' $$ Clearly given any $\epsilon$ we have a positive $\epsilon'$, therefore for any $\epsilon' \geq 0$ it follows that there is $N \in \mathbb{N}^{+}$ such that $$ |\frac{1}{b_n} - \frac{1}{b}| \leq \epsilon' $$ Hence $(1/b_n) \rightarrow 1/b$.

Correction based on yousef magableh answer:

I didn't notice that my $\epsilon'$ was defined in terms of $n$. To avoid that suppose that $b > 0$. Let's pick an $\epsilon = 2b$. Hence that will exists $N_1 \in \mathbb{N}^{+}$ that guarantees that: $$ |b_n-b| \leq 2b \rightarrow -3b \leq -b \leq b_n \leq 3b \rightarrow |b_n| \leq 3b $$

Therefore, for any positive $\epsilon$, there is a positive $\epsilon'$ such that:

$$ \frac{|b_n - b|}{|b_n||b|} \leq \frac{\epsilon}{|b_n||b|} = \frac{\epsilon}{3b|b|} = \epsilon' $$

holds for $n \geq N_1$.

If $b < 0$ just pick $\epsilon = -2b$.

Is it correct? If so, what would you do in order to improve it? Thanks.

Answer 1

I would recommend not assuming anything is positive. It's fine to do that, but is very important to be able to work with absolute values carefully. And absolute value inequalities specifically.

You need to be sure to know when you want a lower vs upper bound. Often the game is bounding something by a series of upper bounds: $<...<...<...$ etc. However to bound something in a denominator, we will need a lower bound on it in order to keep the overall flow of upper bounds. Also we will need such a lower bound (on the thing on the denominator) to be positive, so we stay away from zero.

$$0<a<x<b ~ \text{ implies } ~ 0<\frac 1b < \frac 1x <\frac 1a$$

So a lower bound on $x$ is transformed into an upper bound on $\frac1x$.

Start with $\epsilon=|b|/2$ as already suggested. This will allow us to keep things away from zero.

Then $|b_n-b|<|b|/2$ implies $b-|b|/2<b_n<b+|b|/2$ so $|b_n|>m=\min(|b\pm|b|/2|)$ which we can be certain is positive by the construction. This may seem a bit hard to follow. If so, plug in numbers and try a few examples with specific sequence formula like $1-\frac1n=b_n$ or $-1+\frac1n=b_n$.

Then for any arbitrary $\epsilon$, choose $N$ large enough so that $n>N$ implies both $|b_n-b|<\epsilon m |b|$ and $|b_n-b|< |b|/2$.

Then we have $$ \frac{|b_n-b|}{|b_n||b|}<\frac{\epsilon m |b|}{|b_n||b|}= \frac{\epsilon m}{|b_n|}<\frac{\epsilon m}{m}=\epsilon.$$

Of course there are many ways to do this kind of problem, and this is only one possibility.

Answer 2

No!. The value of N should not depend on $ {b_n} $.

Hint: Try to bound $ \frac{1}{|b_n|} $ (above) by bounding $ {|b_n|} $ (below) using the fact that $ {b_n} $ converges. Let epsilon = $ \frac {|b|}{2} $ in the definition of convergence for $ {b_n} $ . And then you will have a bound on $ {|b_n|} $ .

See:proof (1/bn) -> (1/b) from abbott Understanding Analysis..