5615H: Math Introduction to Analysis I. Homework #10. Problems and Solutions. #1. The sequence {an } is defined by a1 = 0, and an+1 = 2an /2
Fall 2011
for n = 1, 2, 3, . . . .
Prove that {an } is convergent and find its limit. You can use without proof the fact that the function y = f (x) = ax , where a > 0, a = 1, is strictly convex, i.e. f (tx1 + (1 - t)x2 ) < t f (x1 ) + (1 - t) f (x2 ) for all x1 < x2 and 0 < t < 1. Solution. By induction, a1 = 0 < a2 = 1 < a3 < < an < < 2 for all n = 1, 2, 3, . . .. By Theorem 3.14 and its proof, there exists L = lim an = sup{an } 2. Moreover, an+1 = 2an /2 = = an+1 2L/2 , L 2an /2 for n = 1, 2, 3, . . . L 2L/2 , L 2L/2 = L = 2L/2 2.
We claim that L = 2, i.e. the case L = 2L/2 < 2 is impossible. Indeed, if this is the case, then the equation x = 2x/2 has at least 3 distinct solutions x1 = L < x0 = 2 < x2 = 4. In other words, the graph of a strictly convex function y = f (x) := 2x/2 intersects the line y = x at 3 distinct points, which is impossible. Formally, we can write x0 = tx1 + (1 - t)x2 , where t := x2 - x0 x2 - x1 if x0 (x1 , x2 ).
By the property of strict convexity, the desired contradiction follows: x0 = f (x0 ) = f (tx1 + (1 - t)x2 ) < t f (x1 ) + (1 - t) f (x2 ) = tx1 + (1 - t)x2 = x0 . #2. Investigate the behavior (convergence or divergence) of an if n+1- n n (a) an = n + 1 - n; (b) an = ; (c) an = n n - 1 . n (a). Solution. Since the partial sums
n n
sn :=
k=1
ak =
k=1
k+1-
k =
n + 1 - 1 +,
the series an diverges. (b). Note that 0 < an := By Theorem 3.28, the series n+1- n n = n 1 1 < cn := 3/2 . n n+1+ n an also converges.
cn converges. Then by Theorem 3.25(a), 1
(c). We know that n n - 1 0 (Theorem 3.20(c)). Therefore, there exists a natural N0 such that 0 < n n - 1 < 1/2 for all n N0 . Then n 0 < an := n n - 1 2-n for all n N0 , and an converges by Theorems 3.20(e) and 3.25(a).
#3. Using the fact that
n=1
1 2 = , n2 6
find
n=1
(-1)n-1 . n2
Solution. We have
n=1
(-1)n-1 1 1 1 1 = 2 - 2 + 2 - 2 + 2 n 1 2 3 4 = = = 1 1 1 1 1 1 + 2 + 2 + 2 + - 2 + 2 + 2 2 1 2 3 4 2 4 1 1 1 1 1 1 1 + 2 + 2 + - + 2 + 2 + 12 2 3 2 12 2 3 1 1 1 1 2 1- + 2 + 2 + = . 2 12 2 3 12
#4. Find the radius of convergence of each of the following power series: (a) n2 z n , (b) 2n n z , n! (c) 2n n z , n2 (d) n3 n z . 3n
Solution. For each of these series, cn z n , there is a limit := lim(cn+1 /cn ) 0. We can use Theorem 3.34 with an := cn z n , which satisfy
n
lim
an+1 cn+1 z n+1 = lim = |z| . n an cn z n
This implies that the radius of convergence of cn z n is R = 1/. Alternatively, by Theorem 3.37, := lim(cn+1 /cn ) = lim n cn , and the same equality R = 1/ follows from Theorem 3.39. In our particular cases, we have (a) = 1, R = 1; (b) = 0, R = +; 1 (c) = 2, R = ; 2 1 (d) = , R = 3. 3
2