Quote:
Originally Posted by smcdonn2
Real Analysis
Given
a_n, l, and e
we know |a_n-l|<e for convergent sequences
So my question is, to find the right e do we solve the inequality |a_n-l|<e
or a_n<e?
Neither.
There is no 'right e' here. The point is that for *any* choice of e, there exists an N for which the inequality is satisfied by a_N, and all subsequent terms, if the sequence does indeed converge to l.
Perhaps a concrete example would help. Take a simple sequence a_n = 1/n, and we suspect that the limit is 0.
Suppose we choose e arbitrarily to start with, lets say we choose e = 1. Well then, the inequality |a_n-l| < e becomes |1/n| < 1, which is satisfied for n >= 2.
If we chose e = 1/10, then we need |1/n| < 1/10, which is satisfied for n>=11.
For *any* choice of positive e, we need |1/n| < e, which is satisfied for n >= 1/e. So all the terms after a_1/e satisfy the inequality regardless of the choice of e, and so a_n converges to zero.
Last edited by Pyromantha; 09-13-2009 at 06:12 AM.