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I’m tasked with proving the following:

For $n_0 \in \mathbb{N}$ and $q\in \mathbb{R}$ with $0<q<1$ and $n \geq n_0$, let $(a_n)_{n\in\mathbb{N}}$ be a sequence in $\mathbb{R}$. If $$\vert a_{n+1}-a_n\vert\leq q\vert a_n - a_{n-1} \vert$$ then $(a_n)_{n\in\mathbb{N}}$ is a Cauchy sequence.

I thought it may be possible by trying to "form" it in such a way that I can use the ratio test, but it doesn't really work it seems. Alternatively I thought it's sufficient to prove that the sequence converges since that would automatically mean that it's a Cauchy sequence, but it seems to me that there is to little to work with. Any help would be appreciated.

gen-ℤ ready to perish
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Francesco
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As $\forall n>n_o $ $$\left | a_{n+1} -a_n\right | \le q \left | a_{n} -a_{n-1}\right | $$ Let$ n =n_o+1$ $$\left | a_{n_{o}+2} -a_{n_o+1}\right | \le q \left | a_{n_{o}+1} -a_{n_{o}}\right | $$ $n=n_o+2$ $$\left | a_{n_o+3} -a_{n_{o+2}}\right | \le q \left | a_{n_o+2} -a_{n_o+1}\right | \le q^2 \left | a_{n_o+1} -a_{n_{o}}\right | $$ Now generalizing give

$$\left | a_{n+1} -a_{n}\right | \le q^{n-n_o} \left | a_{n_o+1} -a_{n_o}\right | $$ Now let $ \lim_{n \to \infty}$ To get $\lim_{n \to \infty} \left | a_n-a_{n-1} \right | =0$

Aditya Dwivedi
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  • I hope you understand – Aditya Dwivedi Jun 05 '20 at 19:15
  • Thanks for the solution. Two things still remain a bit unclear to me. If we state that $\forall n>n_o$ how are we allowed to say $n =n_o$. Secondly the step from the second to third inequality in the third line is unclear to me. Since 0<q<1 wouldn't $q^2$ make it so that it is less than the inequality with q infront of it? – Francesco Jun 05 '20 at 19:24
  • Thanks for pointing it and we get from 2nd to third by multiplying the first inequality by q – Aditya Dwivedi Jun 05 '20 at 19:33
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    Alright thanks for clarifying. Just to be sure: the last line is the defining criteria to show that it is a Cauchy sequnence, correct? – Francesco Jun 05 '20 at 19:37
  • Unfortunately $|a_{n} - a_{n-1}| \to 0$ does not imply that $a_n$ is a Cauchy sequence. Instead, you should look at the answer mentioned by @rtybase which points to the correct answer. – PAb Jun 05 '20 at 19:54