;line-height:107%;mso-fareast-font-family:\"Times New Roman\"; mso-bidi-font-family:\"Calibri Light\";mso-bidi-theme-font:major-latin; mso-fareast-language:EN-IN\'> Math:

Let sn = 2^1/n+ n sin(np/2), n ? N.

(a) List all subsequential limits of (sn).

(b) Give a formula for nk such that (snk) is an unbounded increasing subsequence of (sn).

(c) Give a formula for nk such that (snk) is a convergent subsequence of (sn).

Sol74:

(a) The subsequential limits of (sn) are 20, 22, and -22. To see this, note that for even n, sn = 21/n + n(1) = 21/n + n, which approaches infinity as n ? 8. For odd n, sn = 21/n - n(1) = 21/n - n, which approaches -8 as n ? 8. Thus, any convergent subsequence of (sn) must consist entirely of even or odd terms. If it consists entirely of even terms, it must converge to 22, since each term is greater than the previous term and approaches infinity. If it consists entirely of odd terms, it must converge to -22, since each term is less than the previous term and approaches -8. Finally, if there is a subsequence with both even and odd terms, it must converge to 20, since every even term is greater than 20 and every odd term is less than 20, and both sequences approach 20 as n ? 8.

(b) Let nk = 2k for all k ? N. Then (snk) = 21/2k + 2k(1) = 21/2k + 2k approaches infinity as k ? 8, since the first term approaches 0 and the second term approaches infinity.

(c) Let nk = 2k-1 for all k ? N. Then (snk) = 21/(2k-1) + (2k-1)(-1) = 21/(2k-1) - (2k-1) approaches a finite limit as k ? 8, since the first term approaches 21 and the second term approaches infinity, but at a slower rate than the first term. Specifically, the limit is -21/2, since the two terms cancel each other out when multiplied.