Sunday, 10 April 2016

sequences and series - Evaluating sqrt6+sqrt6+cdots

Tough as introduction to analysis for beginners (Dutch handbook - I'm Belgian). Again
(n) means index n, x1=6, xn+1=6+xn




  1. Question:




|xn+13|1/5|xn3|



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




  1. Prove that |xn3|(1/5)n1


  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.

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