Round off (beautiful...) Message #1 Posted by Tizedes Csaba on 13 Aug 2003, 1:40 a.m.
In the first math exam we got this example:
What is the limit of this serie: (the calculator use not was allowable!)
an=(1+1/(2^n+1))^(2^(n+1)) where n>infinity (n is integer)
the limit is e^2 (~7.39), but look, what happen if we try to calculate it! Grab your 48, and plot it: 0 40 XRNG 0 10 YRNG 'n' INDEP ERASE DRAW You'll see, the value of function about 37.537.8 will be 1. (not 7.39)
That happen, when 1+1/(2^n+1)=1 for 12 digit, the limit will be 1.
Where? Its simple calculable:
1/(2^n+1)<5e12 > n>37.54
Csaba
