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So my introduction to analysis professor left this problem as an assignment but I just don't know where to start.

Let $\left( a_n\right)_{n\in \mathbb{N}}$ be a sequence of real numbers and $\alpha \in \mathbb{R}$, $\alpha < 1$. Suppose:

$$ \mid a_{n+1} - a_n \mid \leq \alpha \mid a_n - a_{n-1} \mid \forall n\in \mathbb{N}$$ with $2\leq n$

Show that $\left( a_n\right)_{n\in \mathbb{N}}$ converges

Whenever I see ''$n$'', induction comes to my mind, and maybe Cauchy convertion might be the way to follow, but not exactly sure how to proceed on this one.

Any hints?

Thank you

C. Alexander
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    Hint. If you know how to sum a geometric series you can show this sequence is Cauchy. – Ethan Bolker Apr 04 '18 at 19:27
  • A related topic https://math.stackexchange.com/questions/2225451/contraction-mapping-in-the-context-of-fx-n-x-n1/2225503#2225503 – rtybase Apr 04 '18 at 20:33

3 Answers3

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The trick is to apply your inequality many times (formally, by induction), to get that $$ |a_{n+1}-a_n|\leq\alpha^n|a_1-a_0|. $$ Then you can use a geometric series to show that the sequence is Cauchy.

Martin Argerami
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You mentioned induction. Use induction to show $$|a_{n+1} - a_n| \le \alpha^{n-1} |a_2 - a_1|.$$

With this very explicit bound on each increment $|a_{n+1} - a_n|$, you can do a little more work to show this sequence is Cauchy (and thus converges). Specifically use the triangle inequality to write $|a_m - a_n| = |a_m - a_{m-1}| + \cdots + |a_{n+1} - a_n|$, apply the above bound, and use what you know about geometric series to bound $|a_m - a_n|$.

angryavian
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Let $b= |a_1 - a_2| \ge 0$

By induction $|A_N - a_{N+1}| \le \alpha^{N-1}*b$.

If $n,m > N$ then $|a_n - a_m| \le |a_n - a_N| + |a_m - a_M|$.

$|a_n - a_N| \le |a_N - a_{N+1}| + |a_{N+1} - a_{N+2}| + ..... + |a_{n-1} - a_n|\le$

$ b*(\alpha^{N-1} + \alpha^{N} + ...... +\alpha^{n-2})$

$= b\sum\limits_{k=N-1}^{n-2}\alpha^k \le$

$b \sum\limits_{k=N-1}^{\infty}\alpha^k = b(\frac 1{1-\alpha} - \frac {1- \alpha^{N-1}}{1- \alpha} ) = b(\frac {\alpha^{N-1}}{1-\alpha})$.

Likewise $|a_m- a_N| < b(\frac {\alpha^{N-1}}{1-\alpha})$.

So $|a_m - a_n| < 2b(\frac {\alpha^{N-1}}{1-\alpha})$.

If for any $\epsilon > 0$ we can find an $N$ so that $2b(\frac {\alpha^{N-1}}{1-\alpha}) < \epsilon$, we will have proven $\{a_i\}$ is Cauchy.

So let $N > \log_{\alpha} (\frac {(1-\alpha)\epsilon}{2b})$ will be such an $N$.

fleablood
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