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

### 1 Determine the divergence or convergence of ∑ for complex z

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

Therefore ⁄→ 0, implying the divergence of 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

 (1.1)

for all n > N. By separating like so

equation 1.1 informs us that

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 < < 1. Thus the right-hand side of the above inequality converges, implying the convergence of the left-hand side, n=1. 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

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

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

#### (b) ∑zn

This series is a power series of e2z:

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

#### (c) ∑n!zn

Since

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+ for n ≥ 1

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

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

which simultaneously implies that the sequence is bounded and rules out 45 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 .

### 5 If an≥ 0 and ∑an converges, show that ∑ 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 = + + since each an 0. Thus defining xn = 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

converges.

### 6 Find N so that 1 +++++> 100

Define an to be the harmonic series. We have that
Because we know an diverges, we also have that k=02ka2k diverges. So if we can find an integer K such that 1 + k=0K2ka2k > 100 then for N = 2K we’ll have 1 + + + + + > 100. Thus since

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

### 7 Determine whether or not 1 +--++- converges

By grouping the series like so

we see that it is no different than cn where

This is an alternating series for which converges to zero, so our original series must also converge by Rudin’s theorem 3.43.

### 10

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

#### (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) will be I and therefore I - A will be invertibile. Since

we’ll have that I -A is invertible if n=0An 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 = 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 - = 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 n2 , 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.