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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
    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?2012-04-17
  • 0
    @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.2012-04-17
  • 0
    @Auke I think that the second 1. (2. in Auke's comment) should be $|x_n-3|\le(1/5)^{n-1}$.2012-04-17
  • 0
    @Julian, i'd say so too, but that's not what it said :)2012-04-17
  • 0
    Actually I meant <= 1/5 power (n-1) sorry about that.2012-04-17
  • 2
    see also: http://math.stackexchange.com/q/61048/53632012-04-17
  • 2
    see also: http://math.stackexchange.com/questions/115501/on-the-sequence-x-n1-sqrtcx-n2012-04-17

3 Answers 3