Question

Suppose an- is a decreasing sequence of non-negative numbers (that is, 0 S an+1 S an for all n) a) Show that 2K a1 + - n-1 b) Suppose Σ-1 an is a convergent series. Use part a to show that Σ-1 2na2n converges. HINT: recall the monotonic sequence theorem c) Show that n-1 d) Suppose that Ση_1 2na2n is a convergent series. Use part c to show that Ση-1 an

Answer 1

We have a_1 greaterthanorequalto a_1 a_2 greaterthanorequalto a_2 = 1/2 (2a_2) a_3 + a_4 greaterthanorequalto a_4 + a_4 = 2a_4 = 1/2 (4 a_4) since (a_n is downarrow) a_5 + a_6 + a_7 + a_8 greaterthanorequalto a_8 + a_8 + a_8 + a_8 = 4a_8 = 1/2 (8 a_8) (a_2^k -1 + 1 + ..+ a_2k) greaterthanorequalto 1/2 (2^k a_2k) From these inequalitier we get that a_1 + a_2 +.._2k greaterthanorequalto a_1 + 1/2 (2 a_2) + 1/2 (4 a_4) +..+ 1/2 (2^k a_2k) i.e., sigma^2k_n = 1a_n greaterthanorequalto a_1 + 1/2 sigma^k_n = 1 2^n a_2n if sigma^infinity_n = 1a_n is convergent then the partial sum sigma^2k_n = 1 a_n converges to limit l Hence l greaterthanorequalto a_1 + 1/2 sigma^k_n = 1 2^n a_2n Therefore sigma^k_n = 1 2^n a_2n lessthanorequaltoto 2 (l -a_1) so the partial sum for the sigma^infinity_n = 1 2^n a_2n is bounded above and since it is a series of positive terms, it is increasing