Given a sequence s.t $\forall n \in \mathbb{N}, |a_{n+1}-a_n| \leq \frac{9}{10}|a_{n}-a_{n-1}|$ prove the sequence converges.
I tried to use cauchy but it didn't work, looking for a hint.
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paxtibimarce
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2Bounded by geometric sequence – Adam Rubinson Sep 26 '20 at 13:21
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@AdamRubinson this shows that $|a_{n+1}-a_n|$ converges, how do I show it for $a_n$? – paxtibimarce Sep 26 '20 at 13:27
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You misinterpreted Adam Rubinson's comment. $|a_{n+1} - a_n| \to 0.$ – user2661923 Sep 26 '20 at 13:28
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You want a formal proof? I'm guessing something to do with "absolute convergence implies convergence" – Adam Rubinson Sep 26 '20 at 13:31
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I don't get it still. Take $a_n = \sum_{k=1}^n \frac{1}{k}$ we have $|a_{n+1}-a_n| \to 0 $ but it doesn't converge... – paxtibimarce Sep 26 '20 at 13:35
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I don't get your comment above. – Adam Rubinson Sep 26 '20 at 13:36
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@AdamRubinson what you have showed as far as I understand is that $|a_{n+1}-a_n|$ converges to 0. my example shows that is not enough for convergence of $a_n$ – paxtibimarce Sep 26 '20 at 13:38
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No your sequence $(1/n)$ doesn't satisfy the inequality for all $n. (\frac{1000}{1001} - \frac{999}{1000}) $ is not $\leq \frac{9}{10}(\frac{999}{1000} -\frac{998}{999})$ – Adam Rubinson Sep 26 '20 at 13:49
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Look at this answer. Your answer is there, within the lines ... – rtybase Sep 26 '20 at 14:01
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"I don't get it still. Take ...". That's a series that doesn't converge; you asked about a sequence. Although "sequence" and "series" are almost interchangeable in daily English, they are distinctly different things in math (although they share some notation). – JonathanZ Sep 26 '20 at 14:27
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Edited my answer to actually answer the question.
I'll be a bit more explicit than akrihnab, but the same proof strategy.
$\forall i,j \in \mathbb{N}$ with $i>j,$
$$|a_i - a_j| \leq \left( \ \left(\frac{9}{10}\right)^{i-1} + ... + \left(\frac{9}{10}\right)^{j}\ \right)|a_1-a_0| = \left(\frac{1-(9/10)^{i}}{1-9/10} - \frac{1-(9/10)^{j}}{1-9/10}\right)|a_1-a_0| = 10 \left(\left(\frac{9}{10}\right)^{j}-\left(\frac{9}{10}\right)^{i}\right)|a_1-a_0| \leq 10\left(\frac{9}{10}\right)^{j}|a_1-a_0| .$$
Conclusion of above working: $\ \forall n,m \in \mathbb{N}, \ |a_n - a_m| \leq 10|a_1-a_0|\left(\frac{9}{10}\right)^{\min{(n,m)}}.$
Let $\epsilon > 0.$ Then $\exists \ N_0$ such that $\left(\frac{9}{10}\right)^{N_0} < \frac{\epsilon}{10|a_1-a_0|}.\ $ So for $N_0 > \log_{(10/9)}\left(\frac{10|a_1-a_0|}{\epsilon}\right) = \log_{(9/10)}\left(\frac{\epsilon}{10|a_1-a_0|}\right) $:
$\epsilon > 10|a_1-a_0|\left(\frac{9}{10}\right)^{N_0} \geq |a_n-a_m| \ \forall n,m \geq N_0 \implies$ Sequence is Cauchy.
Adam Rubinson
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ok I think i get you now, so you showed that $|a_n - a_0|$ converges here right? but i'm not sure about $a_n$ yet though it does seem closer. – paxtibimarce Sep 26 '20 at 13:40
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$|a_n-a_0|$ does converge (which is the same as showing that $a_n$ converges, since $a_0$ is just a fixed real number because we are given the sequence). However I haven't proven this yet: all I have shown is that $a_n$ is bounded because $|a_n-a_0|$ is bounded above by $10|a_1-a_0|$, so I haven't quite answered your question yet. I'm sure we should be able to show that the sequence is Cauchy... – Adam Rubinson Sep 26 '20 at 13:55
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Hmm I have to go out now which is a shame. Will come back to try to answer this later. – Adam Rubinson Sep 26 '20 at 14:05
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We show that the sequence is Cauchy.
Let $|a_m - a_n|$ with $m > n$. Then $$|a_m - a_n | \leq |a_{m}- a_{m-1}| + \cdots |a_{n+1} - a_n |$$ by triangular inequality. As stated in one of the comments $|a_{k+1} - a_{k}| <= \frac{9^k}{10^k}|a_1 - a_0|$ for any $k$. Since the geometric series $\sum_{k \in \mathbb{N}} \frac{9^k}{10^k}$ converges we conclude that the sequence is Cauchy.
Infinity_hunter
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