Math 508: Advanced Analysis

Homework 5
Lawrence Tyler Rush
<me@tylerlogic.com>
October 7, 2014
http://coursework.tylerlogic.com/courses/upenn/math508/homework05

1 Determine the divergence or convergence of ∑ --1n- 1+z for complex z

If |z|≤ 1 the |zn|≤ 1 which implies |1 + zn|≤ 2 and subsequently that
||--1--|| 1 ||1+ zn|| ≥ 2

Therefore 11+zn- ⁄→ 0, implying the divergence of ∑ 11+zn in this case.

Now assume that |z| > 1. Then |zn| increases as n gets large and |zn + 1| > 2. So let N be such that |zn| > 2 for all n > N. Then

|zn + 1| ≥ |zn| > |zn|- 1 ≥ |zn|- |zn|= |z|n 2 2
(1.1)

for all n > N. By separating ∑ |1+1zn| like so

∞∑ ---1--- ∑N --1---- ∑∞ --1---- |1+ zn| = |1 + zn | + |1+ zn| n=1 n=1 n=N+1

equation 1.1 informs us that

∞∑ 1 ∑N 1 ∞∑ 2 ∑N 1 ∞∑ ( n√2-)n -----n-< -----n-+ --n-= -----n-+ --- n=1 |1 + z | n=1|1+ z | n=N+1 |z| n=1|1+ z | n=N+1 |z|

Now the left addend of the right-hand side of the above inequality is a finite sum and the right addend is a convergent geometric series since 0 < n√- -|z2| < 1. Thus the right-hand side of the above inequality converges, implying the convergence of the left-hand side, ∑ n=1∞|1+1zn|. This in turn implies that our series in question is absolutely convergent when |z| > 1, and therefore convergent.

2 If an > 0, ∑ an converges, and {bn} is bounded show that ∑a nbn converges.

Let ∑ an be a convergent series with each an > 0 and {bn} a bounded sequence. Define M as the bound on {bn} so that |bn| < M for all n. Then we have
∑ ∑ ∑ ∑ |anbn| = an|bn| < anM = M an

so due to the convergence of ∑ an, then M ∑ an and therefore ∑ |anbn| converges. Hence ∑ anbn converges absolutely, and so converges.

3 Find the radius of convergence of the following power series.

(a) ∑ n3zn

Since
||(n+ 1)3zn+1|| ||n3 + 3n2 + 3n+ 1 || ||-----33----|| = ||-------3------z|| n z n

approaches z as n increases, then the ratio test tells us that the radius of convergence is 1.

(b) ∑ 2n n!zn

This series is a power series of e2z:
∑ 2n ∑ 2zn --zn = --- = e2z n! n!

and so this series converges for all z, i.e. the radius is infinite.

(c) ∑ n!zn

Since
||(n+ 1)!zn+1 || ||---n!z3----|| = |(n + 1)z|

approaches ∞ as n increases, then the ratio test tells us that the radius of convergence is 0, but technically, z = 0 will make the series converge.

4 Detemine the limit of {an} where a0 = 0 and an+1 = an2 + 245 for n ≥ 1

We first note that if the limit is finite, say L, then it must satisfy L2 - L + 425- = 0 given the recursive definition. Therefore, L can only be 15 or 45 since these are the roots of that equation.

Now since a0 = 0, the recusive formula just adds a positive value of 425, informing us that this sequence if monotonically increasing. Furthermore, if an < 1 5 we see that

4 ( 1)2 4 1 an+1 = a2n +--< - + -- = - 25 5 25 5

which simultaneously implies that the sequence is bounded and rules out 4∕5 as a possible limit. Thus the sequence is monotonically increasing and bounded, implying that it must converge, and because the initial value is zero, the only point it can converge to is 1 5.

5 If an ≥ 0 and ∑ an converges, show that ∑ √ --- -nan converges

Let an ≥ 0 and assume ∑ an converges. Then the partial sums {sn} with sn = a1 + ⋅⋅⋅ + an converge. This implies the convergence of {tn} where tn = √a1- + ⋅⋅⋅ + √an- since each an ≥ 0. Thus defining ∑ xn = ∑ √an- implies that {tn} is the sequence of partial sums of ∑ xn and they form a bounded sequence. Furthermore, defining ∑ yn as the harmonic series, we have y0 ≥ y1 ≥ y2 ≥⋅⋅⋅ and limn→∞yn = 0. Thus Rudin’s Theorem 3.42 informs us that the series
√ --- ∑ x y = ∑ --an n n n

converges.

6 Find N so that 1 + 1 2 + 1 3 + 1 4 + ⋅⋅⋅ + 1- N > 100

Define ∑ an to be the harmonic series. We have that
( ) ( ) ( ) ∑ a = 1 + 1 + 1 + 1 + 1 + 1 + 1+ 1 + ⋅⋅⋅ n 2 3 4 5 6 7 8 (1 ) ( 1 1) (1 1 1 1 ) > 1 + 2 + 4 + 4 + 8 + 8 + 8 + 8 + ⋅⋅⋅ ∞∑ = 1 + 2k-1-- k=0 2k+1 ∑∞ = 1 + 1 2k 1k 2k=0 2 1∑∞ = 1 + 2 2ka2k k=0
Because we know ∑ an diverges, we also have that ∑ k=0∞2ka2k diverges. So if we can find an integer K such that 1 + 1 2 ∑ k=0K2ka2k > 100 then for N = 2K we’ll have 1 + 1 2 + 1 3 + 1 4 + ⋅⋅⋅ + 1- N > 100. Thus since
K∑ ∑K 1+ 1 2ka2k = 1+ 1 2k-1k = 1+ K- 2 k=0 2 k=0 2 2

then 1 + K2- > 100 implies that K > 198, so K = 199 and therefore N = 2199 will satisfy.

7 Determine whether or not 1 + 1 2 -1 3 -1 4 + 1 5 + 1 6 -⋅⋅⋅ converges

By grouping the series like so
( 1) (1 1) ( 1 1 ) ( 1 1) 1 + 2 - 3 + 4 + 5 + 6 - 7 + 8 + ⋅⋅⋅

we see that it is no different than ∑ cn where

( ) n+1 --1--- -1- cn = (- 1) 2n - 1 + 2n

This is an alternating series for which ( ) 2n1-1 + 12n converges to zero, so our original series must also converge by Rudin’s theorem 3.43.

8

(a)

(b)

9

10

Let A be an n × n matrix, and use the norm of homoework 3 problem 6.

(a) Compute (I - A)(I + A + A2 + ⋅⋅⋅ + AN)

(I - A)(I + A + A2 + ⋅⋅⋅+ AN ) = I - AN+1

(b) Show that if |A| < 1, then I - A is invertible.

Let |A| < 1. Then as N gets large, I - AN+1 approaches I, so in light of the previous part of this problem, if (I -A)∑ n=0NAn converges as N gets large, then (I -A)∑ ( An) will be I and therefore I - A will be invertibile. Since
∑N n ∑N n ∑∞ n Nli→m∞(I - A) A = (I - A )Nli→m∞ A = (I - A) A n=0 n=0 n=0

we’ll have that I -A is invertible if ∑ n=0∞An converges. Thus because ∑ |An|≤∑ |A|n, the right side converges since it’s a geometric series and |A| < 1. This tells us that ∑ An converges absolutely and that therefore ∑ An converges. Hence we have our desired result of I - A being invertible.

(c) Show the set of invertible matrices is open

Let ε = -1-- |A-1| and B be a matrix in the ε-ball of A. Then we have that |A-B| < ε implying that |A-1||A-B| < 1. Thus |I -A-1B| < 1. By the previous part of the problem, we then know that I -( - 1 ) I - A B = A-1B is invertible. Then there is some invertible matrix C with CA-1B = I. But then B is invertible with inverse CA-1. Hence the ε-ball of A contains only invertible elements, and so the set of invertible matrices is open.

11 Prove that the set of othogonal n × n matrices are compact.

Since n×n matrices are a subspace of ℝn2 , then to show this set is compact, we need only show that it’s closed and bounded.

Bounded. Since any othogonal matrix A has that AAt = I, then with our definition of the norm, |A|2 = ⟨A,A⟩ = trace(AAt) = trace(I) = n. Thus the set of othogonal matrices is bounded.

Closed.

12

(a)

(b)