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$|a_{n+1}-a_n| \leq \frac{1}{2} |a_{n}-a_{n-1}|$. I'm trying to show that $(a_n)$ is convergent by showing that $(a_n)$ is Cauchy.

let $|a_2 - a_1 | =x$.

$|a_3-a_2| \leq \frac{1}{2}x$

$|a_4-a_3| \leq \frac{1}{2}|a_3-a_2|\leq \frac{1}{4}x$

In general, for every $n$:

$|a_{n+1}-a_n| \le \frac {1}{2^{n-1}}x$

Let $\epsilon >0$

for every $m,n > 0 $, say $m>n$ the following holds:

$$|a_m-a_n| = |a_m-a_{m-1} + a_{m-1} - a_{m-2} \dots a_{n+1}- a_n| \le$$

$$|a_m-a_{m-1}| + |a_{m-1} - a_{m-2}| + \dots |a_{n+1}-a_n| \leq \frac{x}{2^{m-2}} + \frac{x}{2^{m-3}} + \dots + \frac{x }{2^{n-1}}$$.

I can say that for big enough $n,m$, $|a_m-a_n|<\epsilon$ and therefore, this sequence is Cauchy? It sounds not good enough to me yet I don't know what else to do here.

1 Answers1

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Your first steps are correct. To finish

$$|a_m - a_n| \leqslant \frac{1}{2^{m-2}} + \frac{1}{2^{m-3}} + \dots + \frac{1}{2^{n-1}} \\ = \frac{1}{2^{n-1}} \sum_{k=0}^{m-n-1}\frac{1}{2^k} \ < \frac{1}{2^{n-1}} \sum_{k=0}^{\infty}\frac{1}{2^k} = \frac{1}{2^{n-1}}\frac{1}{1- 1/2} = \frac{1}{2^{n-2}} .$$

Since $1/2^{n-2} \to 0$ as $n \to \infty$ it follows that for any $\epsilon > 0$ there exists $N$ such that if $n \geqslant N$ and for any $m > n$, then $|a_m - a_n| < \epsilon$ and the sequence is Cauchy.

RRL
  • 90,707
  • how can you say for sure that when this expression limit is 0, $m\geq n$? –  Feb 27 '17 at 20:25
  • Are you asking why $1/2^{n-2}$ converges to $0$ or how $m$ figures into this? Regarding $m$, the inequality is true for any $m > n$ -- we find an upper bound in $1/2^{n-2}$ that is nicely independent of $m$. So if $1/2^{n-2}$ becomes small the Cauchy criterion must be satisfied. – RRL Feb 27 '17 at 20:35
  • S. Peter: You're welcome. – RRL Feb 27 '17 at 20:38