Consider the following sequence: As we progress into this sequence we encounter more and more zeroes while 1s get sparser. Prove that this sequence does NOT converge to 0. Hint: try to determine the values of epsilon > 0 for which it possible to find a suitable N , and those epsilon > 0 for which no such N exists. Solution Notice that 1 repeats itself after 1,2,3... \'0\' After some n\'th number,1 re-occurs after \'m\' zeros. Notice m < n Therefore, at n\'th number, 1 will occur before or at (n+m) th term, or simply before (n+n) term = 2n term Thus we can say if n-term is zero, 1 will occur before 2n term. just let epsilon = 1/2 We now want \'N\' , how so ever large, so that , after n>N, the sequence has value less than 1/2 But how so ever large that N may be, atleast one \'1\' will occur between N and 2N, and 1 is not less than epsilon (=1/2) Therefore there is no such N for given epsilon, to satisfy convergence. .

1. Consider the following sequence: As we progress into this sequence we encounter more and
more zeroes while 1s get sparser. Prove that this sequence does NOT converge to 0. Hint: try to
determine the values of epsilon > 0 for which it possible to find a suitable N , and those epsilon >
0 for which no such N exists.
Solution
Notice that 1 repeats itself after 1,2,3... '0' After some n'th number,1 re-occurs after 'm' zeros.
Notice m < n Therefore, at n'th number, 1 will occur before or at (n+m) th term, or simply
before (n+n) term = 2n term Thus we can say if n-term is zero, 1 will occur before 2n term. just
let epsilon = 1/2 We now want 'N' , how so ever large, so that , after n>N, the sequence has
value less than 1/2 But how so ever large that N may be, atleast one '1' will occur between N
and 2N, and 1 is not less than epsilon (=1/2) Therefore there is no such N for given epsilon, to
satisfy convergence.