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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

Bernard
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  • First try to find out, why $b_k$ must tend to $2$, then apply the proof. – Peter Sep 14 '20 at 20:26
  • There are no series, just a sequence. It is clear that if $(a_k)$ converges to $1/2$, then $(b_k)$ converges also. So it is Cauchy. Your exercise is to prove that using the definition of Cauchy sequences. – TheSilverDoe Sep 14 '20 at 20:26
  • Yes but im struggling as to prove it using the definition, any advice with how to prove it using the definition would be appreciated –  Sep 14 '20 at 20:45
  • $b_k$ converges to $1/(1-1/2)=2$ and so it is Cauchy (every convergent sequence is Cauchy) – Mittens Sep 14 '20 at 20:45
  • If a sequence $x_n$ converges to $x$, then for any $\varepsilon>0$ there is $N_\varepsilon$ such that $|x-x_n|<\varepsilon/2$ for all $n\geq N_\varepsilon$, Then, for all $n,m\geq N_\varepsilon$ $$|x_n-x_m|\leq|x_n-x|+|x-x_m|<\varepsilon/2 +\varepsilon.2=\varepsilon$$. – Mittens Sep 14 '20 at 20:47
  • apologies i left off the fact that they told us to pretend we do not know that bk is convergent in the question –  Sep 14 '20 at 20:48

3 Answers3

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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.

Jingeon An-Lacroix
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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?

Bernard
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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 $n,m\geq N$.

Mittens
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