Analysis Homework #7 Solutions 1. Define a sequence {an } by letting a1 = 2 and an+1 = 4 − 1 an for each n ≥ 1. Show that 2 ≤ an ≤ an+1 ≤ 4 for each n ≥ 1 and find the limit of this sequence. • Since the first two terms are a1 = 2 and a2 = 4 − 1/2 = 7/2, the statement 2 ≤ an ≤ an+1 ≤ 4 does hold when n = 1. Suppose that it holds for some n, in which case 1 1 1 1 ≥ ≥ ≥ 2 an an+1 4 =⇒ =⇒ 1 1 1 1 ≤4− ≤4− ≤4− 2 an an+1 4 2 ≤ an+1 ≤ an+2 ≤ 4. 4− Thus, the statement holds for n + 1 as well, so it actually holds for all n ∈ N. This shows that the given sequence is monotonic and bounded, hence also convergent; let us denote its limit by L. Then the definition of the sequence gives an+1 = 4 − 1 an =⇒ L=4− 1 L =⇒ L2 − 4L + 1 = 0, so we can solve this quadratic equation to get √ √ √ 4 ± 16 − 4 4±2 3 L= = = 2 ± 3. 2 2 Since an ≥ 2 for all n, however, we must have L ≥ 2 and this implies L = 2 + √ 3. 2. Suppose {an } is a sequence of non-negative terms. Show that its sequence of partial sums sn = a1 + a2 + . . . + an converges if and only if sn is bounded for all n. • If sn converges, then sn is bounded by Theorem 5.7 in your notes. • If sn is bounded, on the other hand, then sn is monotonic and bounded because sn+1 = a1 + a2 + . . . + an + an+1 = sn + an+1 ≥ sn . Being monotonic and bounded, sn must converge by Theorem 5.8 in your notes. 3. Define a sequence {an } by letting a1 = 1 and an+1 = na2n n+1 for each n ≥ 1. Show that 0 ≤ an+1 ≤ an ≤ 1 for each n ≥ 1 and find the limit of this sequence. 2 nan • It is clear that an+1 = n+1 ≥ 0 for each n. Moreover, a1 ≤ 1 and if we assume that an ≤ 1 for some n, then we immediately get na2n ≤ a2n ≤ 1 n+1 an+1 = as well. This implies that an ≤ 1 for all n and it also implies that an+1 = na2n ≤ a2n ≤ an n+1 for all n. Thus, the given sequence is monotonic and bounded, hence also convergent. If we denote its limit by L, then we must have an+1 = na2n n+1 =⇒ nL2 = L2 n→∞ n + 1 L = lim =⇒ L = 0, 1. Since {an } is decreasing with a2 = 1/2, however, we have L ≤ 1/2 and so L = 0. ∑ ∑∞ 1 4. Suppose the series ∞ n=1 an converges. Show that the series n=1 1+an diverges. ∑ • Since ∞ n=1 an converges, we have limn→∞ an = 0 by the nth term test, so lim n→∞ 1 = 1. 1 + an Using the nth term test once again, we conclude that ∑∞ 1 n=1 1+an diverges.