Example 5 Determine the values of x for which the sequence

1. Example 5 Determine the values of x for which the sequence for n>0, is bounded below or bounded above. Solution Case I:0  x  1 In Example E5.2A (5), we showed this sequence is decreasing. Therefore it is bounded above by a1(x)=x. It is bounded below by 0. Since the sequence is bounded above and and below, it is bounded. Case II:x > 1 In Example 5.2A (5) we showed that for n  1: For x > 1 the sequence is bounded below by 0. We show this sequence is never bounded above. FixN > 1/(x-1). Then N(x-1) > 1 and Nx > N+1.Let p  N. By the above computation:

2. for p N (*) where are fixed numberswhich do not depend on p. By (*): ap+1(x)  kap(x)for p N. Apply this formula q times: for q  1. Since k>1, the sequence {kqA} is unbounded. Hence the sequence {an(x)} is also unbounded. Case III:x < -1 The even terms of this sequence are positive while the odd terms are negative. By Case II, the even terms are not bounded above and the absolute values of the odd terms are not bounded above. Therefore the odd terms, being negative, are not bounded below. Hence this sequence is neither bounded above nor bounded below. Case IV:–1  x < 0 By Case I, the |an(x)| lie between 0 and 1. Hence the an(x) lie between –1 and 1. Thus the sequence {an(x)} is bounded. Summary This sequence {an(x)} is bounded below for –1  x and is bounded above for –1  x  1.