If $a_k$ is a sequence that converges to $1/2$, prove that $b_k = 1/(1-a_k)$ is Cauchy

Question

The question asks to prove directly from the definition of a Cauchy sequence that $b_k$ is Cauchy, but I am hopelessly confused, these are evidently series approaching infinity

Answer 1

This may seem as cheating. But: $f(x)=1/(1-x)$ is continuous everywhere except on $x=1$. Therefore since $a_k$ converges to $1/2$, $$\lim_{k\rightarrow \infty}b_k=\lim_{k\rightarrow\infty}\frac{1}{1-a_k}=\frac{1}{1-\lim_{k\rightarrow\infty}a_k}=\frac{1}{1-1/2}=2.$$ The claim $b_k$ is Cauchy trivially follows from the fact that $b_k$ converges.

Yet, I'll call this as a rigorous proof since you can prove the continuity easily.

Answer 2

Hint: To prove directly that $(b_k)$ is Cauchy, compute first $$b_k-b_\ell=\frac1{1-a_k}-\frac 1{1-a_\ell}=\frac{a_k-a_\ell} {(1-a_k)(1-a_\ell)}$$ and note that, since $(a_k)$ converges to $\dfrac12$, if $k,\ell$ are large enough, we have, say $$1-a_k,\enspace1-a_\ell>\frac14.$$ Can you take it from there?

Answer 3

Hint:

Since $a_n\rightarrow\frac12$, there is $N$ such that $|a_n-1|>\frac{1}{3}$ for all $n\geq N. (details left to you)

$\Big|\frac{1}{1-a_n}-\frac{1}{1-a_m}\Big|=\frac{|a_n-a_m|}{|1-a_n|\,|1-a_m|}\leq 9|a_n-b_m|$

As $a_n$ converges, $\{a_n\}$ is Cauchy and so, given $\varepsilon>0$, there is $N_2\geq N$ such that $|a_n-a_m|<\varepsilon/9$ for all $n,m\geq N$.