Tough as introduction to analysis for beginners (Dutch handbook - I'm Belgian). Again ($n$) means index $n$, $x_1 = \sqrt6$, $x_{n+1} = \sqrt{6+x_n}$

  1. Question:

$$|x_{n+1} - 3| \le 1/5 \cdot |x_n - 3|$$

For me this means that $3$ as a 'limit', we need to find that the distance between $x_{n+1}$ and the 'limit' is $1/5$ the distance between the $x_n$ and the limit. Where does the $1/5$ come from?

  1. Prove that $|x_n - 3|\le (1/5)^{n-1}$

  2. prove that the sequence converges to $3$.

ps: When I studied maths in 1980. we went quickly towards metric spaces, so these calculus minded times are nothing compared to those times. But still, as I didn't pass then, I'd like to restart on a new basis. Thanks for all the help. If you know where maths can be studied in community on the net, always welcome.

  • 1
    $\begingroup$ The questions asked are: 1. prove that $|x_{n+1}-3|\leq \frac{1}{5}|x_n-3|$. 2. prove that$|x_n-3|\leq \frac{1}{5}\exp(n-1)$. 3. conclude that $x_n$ converges to 3. Have I got this right? $\endgroup$ – akkkk Apr 17 '12 at 13:04
  • $\begingroup$ @Ignace, welcome to the site! I took the liberty of TeXifying your question, 'veterans' often do this :-). Please check that I didn't screw up your intended notation. You are also welcome to look at the modifications to learn a few things about TeX (for future use). I also replaced the tags with ones that I think fit a bit better. $\endgroup$ – Jyrki Lahtonen Apr 17 '12 at 13:14
  • $\begingroup$ @Auke I think that the second 1. (2. in Auke's comment) should be $|x_n-3|\le(1/5)^{n-1}$. $\endgroup$ – Julián Aguirre Apr 17 '12 at 13:19
  • 2
    $\begingroup$ see also: math.stackexchange.com/q/61048/5363 $\endgroup$ – t.b. Apr 17 '12 at 13:54
  • 2
    $\begingroup$ see also: math.stackexchange.com/questions/115501/… $\endgroup$ – Aryabhata Apr 17 '12 at 15:41

For the inequality, by the definition of $x_n$ we have
$$x_n-3=\sqrt{6+x_{n-1}}-3.$$ Multiply by $\dfrac{\sqrt{6+x_{n-1}}+3}{\sqrt{6+x_{n-1}}+3}$. So we are multiplying by $1$ in a fancy way. We get $$x_n-3=\frac{x_{n-1}-3}{\sqrt{6+x_{n-1}}+3}.\tag{$\ast$}$$ The bottom is clearly $>5$, since the $x_i$ start and stay positive. One can do better than $5$ here, for example we can without thought replace $5$ by $\sqrt{6}+3$, and with not much more by $\sqrt{6+\sqrt{6}}+3$. But it doesn't matter, $5$ is good enough for a proof of convergence. It would even be enough to observe that the denominator is $>3$. Taking absolute values, we find that $$|x_n-3|=\frac{|x_{n-1}-3|}{\sqrt{6+x_{n-1}}+3}<\frac{|x_{n-1}-3|}{5}.$$ Iterate. The distance to $3$ gets divided by at least $5$ with each iteration, so after a (short) while $x_n$ is awfully close to $3$. Thus our sequence has limit $3$.

  • $\begingroup$ Nice answer..+1 $\endgroup$ – IDOK Jun 20 '12 at 7:32

Let's define the auxiliary sequence $a_{(n)}$, $n\ge1$, n in${\mathbb N}$ as follow:

$$a_{n}=\frac{|x_{n+1} - 3|}{|x_n - 3|}$$

i). Taking into account that $x_{n}$ is positive, one sees that $|\sqrt{6+x_n}+3|>5$. Hence, our first inequality may be proved as follows:

$$a_{n}=\frac{|\sqrt{6+x_n}-3|}{|x_n - 3|}=\frac{1}{|\sqrt{6+x_n}+3|}\le \frac{1}{5} \to \space a_{n}\le \frac{1}{5}. $$

ii). Proving the second inequality: $$a_{1}\cdot a_{2} \cdot a_{3}\cdots a_{n-1}=\frac{|x_{n} - 3|}{|\sqrt6 - 3|}\le \left({\frac{1}{5}}\right)^{n-1} \to \space |x_{n} - 3|\le {|\sqrt6 - 3|}\left({\frac{1}{5}}\right)^{n-1} \le \left({\frac{1}{5}}\right)^{n-1}.$$

iii). Using the inequality from the previous point we get immediately that:

$$\lim_{n\to\infty} |x_{n} - 3|\le 0 \to \lim_{n\to\infty} x_{n}=3.$$

The proof is complete.

  • $\begingroup$ In 1), you seem to use $\to$ as a reverse implication sign... Is this so? $\endgroup$ – Did Jun 20 '12 at 6:10
  • $\begingroup$ Precisely: involves $\ne$ implies. $\endgroup$ – Did Jun 20 '12 at 6:19
  • $\begingroup$ @did: i did change it. $\endgroup$ – user 1357113 Jun 20 '12 at 6:23
  • $\begingroup$ Sorry but the formulation is still very odd. Why not doing it in the proper order? (i) Define $a_n$. (ii) Manipulate the formula for $a_n$ and add a (missing at present) argument to prove that $a_n\leqslant1/5$. $\endgroup$ – Did Jun 20 '12 at 6:53
  • $\begingroup$ @did: hehe. Right. I just improved that! $\endgroup$ – user 1357113 Jun 20 '12 at 7:14

We will also assume that $x_n$ lies between 0 and 3.

$3-x_{n+1} = 3 - \sqrt{6 + x_n} = 3 - \sqrt{9-(3-x_n}$.

Call $3-x_n$ as $a_n$.

So $a_{n+1} = 3 - \sqrt{9-a_n} = 3(1 - \sqrt{1-\frac{a_n}{9}}) \le 3\frac{a_n}{18} = \frac{a_n}{6}$ (by the Taylor series expansion).

Also, $a_{n+1}$ does not become negative like this, and as we will show its absolute value keeps decreasing. So we justified in assuming that $x_n$ lies between 0 and 3.

So $a_{n+1} \le \frac{a_n}{5}$ also

As $a_{n+1} \le \frac{a_n}{5}$, and $a_1$ = $3 - \sqrt{6} \lt 1$, $a_n \le \frac{1}{5}e^{n-1}$ in general.

So the series converges to 3.

  • $\begingroup$ Sorry but $a_{n+1}\leqslant\frac16a_n$ is wrong. $\endgroup$ – Did Jun 20 '12 at 7:46

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