October 7, 2014

http://coursework.tylerlogic.com/courses/upenn/math508/homework05
If |z|≤ 1 the |z

Therefore ⁄→ 0, implying the divergence of ∑ in this case.

Now assume that |z| > 1. Then |z^{n}| increases as n gets large and |z^{n} + 1| > 2. So let N be such that |z^{n}| > 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.

Let ∑ a

so due to the convergence of ∑
a_{n}, then M ∑
a_{n} and therefore ∑
|a_{n}b_{n}| converges. Hence ∑
a_{n}b_{n} converges absolutely,
and so converges.

Since

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

This series is a power series of e

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

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.

We first note that if the limit is finite, say L, then it must satisfy L

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

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 .

Let a

converges.

Define ∑ a

then 1 + > 100 implies that K > 198, so K = 199 and therefore N = 2^{199} will satisfy.

By grouping the series like so

we see that it is no different than ∑
c_{n} where

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

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

Let |A| < 1. Then as N gets large, I - A

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

Let ε = and B be a matrix in the ε-ball of A. Then we have that |A-B| < ε implying that |A

Since n×n matrices are a subspace of ℝ

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