NoteNextra-origin/content/Math4111/Math4111_L21.md

Lecture 21

Review

Recall the alternating series test from calculus: "Suppose (a_n)^\infty_{n=1} is a sequence satisfies the following conditions:

1. The sequence is nonnegative. (For all n\in \mathbb{N}, a_n\geq 0.)
2. The sequence is decreasing. (a_1\geq a_2\geq a_3\geq \cdots)
3. \lim_{n\to\infty}a_n=0.

Then \sum_{n=1}^\infty (-1)^{n+1}a_n converges."

Exercise: Show that the statement above is false if we remove the second condition.

[Hint: Use the fact that \sum_{n=1}^\infty \frac{1}{n} diverges.]

Let the sequence a_n be defined as a_n=\frac{1}{n},a_{n+1}=0 for all n\in \mathbb{N}. This sequence satisfies the 1,3 but not the 2.

And the harmonic series is not convergent.

New Material

Other tests for convergence of series

Recall the integration by parts formula:

Let A(t),a(t),b(t) be functions of t and A'(t)=a(t).

Then

\begin{aligned}
\int_p^q a(t)b(t)\,dt&=\int_p^q b(t)A'(t)\,dt\\
&=\left.b(t)A(t)\right|_p^q-\int_p^q A(t)b'(t)\,dt
\end{aligned}

Theorem 3.41 Summation by parts

Let a_n,b_n be sequences.

Let A(n)=\sum_{k=1}^n a_k. (A_{-1}=0). If 0\leq p\leq q, then

\sum_{n=p}^q a_nb_n=A_q b_q-A_{p-1}b_p-\sum_{n=p}^{q-1}A_n (b_n-b_{n+1})

Proof:

\begin{aligned}
\sum_{n=p}^q a_nb_n&=\sum_{n=p}^q (A_n-A_{n-1})b_n\\
&=\sum_{n=p}^q A_nb_n-\sum_{n=p}^q A_{n-1}b_n\\
&=\sum_{n=p}^q A_nb_n-\sum_{n=p-1}^{q-1}A_n b_{n+1}\\
&=A_qb_q-A_{p-1}b_p-\sum_{n=p}^{q-1} A_nb_n-\sum_{n=p}^{q-1} A_n b_{n+1}\\
&=A_qb_q-A_{p-1}b_p-\sum_{n=p}^{q-1} A_n (b_n-b_{n+1})
\end{aligned}

QED

Theorem 3.42 (Dirichlet's test)

Suppose

(a) the partial sum A_n of \sum a_n form a bounded sequence.
(b) b_0\geq b_1\geq b_2\geq \cdots (non-increasing)
(c) \lim_{n\to\infty}b_n=0.

Then \sum a_nb_n converges.

Proof:

By Cauchy criterion, it's enough to prove

\forall \epsilon >0, \exists N\in \mathbb{N} such that for all p\geq q\geq N,

\left|\sum_{n=p}^q a_nb_n\right|<\epsilon

By the partial sum A_n of \sum a_n form a bounded sequence. Let \left|A_n\right|\leq M for all n\in \mathbb{N}.

\begin{aligned}
\left|\sum_{n=p}^q a_nb_n\right|&=\left|A_qb_q-A_{p-1}b_p-\sum_{n=p}^{q-1}A_n (b_n-b_{n+1})\right|\\
&\leq |A_qb_q|+|A_{p-1}b_p|+\sum_{n=p}^{q-1}|A_n (b_n-b_{n+1})|\\
&\leq M|b_q|+M|b_p|+\sum_{n=p}^{q-1}M(b_n-b_{n+1})\\
&=M|b_q|+M|b_p|+M\sum_{n=p}^{q-1}(b_n-b_{n+1})\\
&=M|b_q|+M|b_p|+M(b_p-b_q)\\
&=2M|b_p|
\end{aligned}

Then we let \epsilon >0 be given. Since b_n\to 0, there exists N\in \mathbb{N} such that for all n\geq N, |b_n|<\frac{\epsilon}{2M}.

If q\geq p\geq N, then

\left|\sum_{n=p}^q a_nb_n\right|\leq 2M|b_p|<\epsilon

So \sum a_nb_n converges.

QED

Theorem 3.43 (Alternating series test)

Let (b_n)^\infty_{n=1} be a sequence such that:

(a) b_1\geq b_2\geq b_3\geq \cdots (non-increasing) (b) \lim_{n\to\infty}b_n=0

Then \sum_{n=1}^\infty (-1)^{n+1}b_n converges.

Proof:

Let a_n=(-1)^{n+1}

A_n=\sum_{k=1}^n a_k=1 if n is odd, 0 if n is even.

So |A_n|\leq 1 for all n\in \mathbb{N}.

By Theorem 3.42, \sum_{n=1}^\infty a_n b_n converges.

QED

Example:

Consider the power series \sum_{n=0}^\infty \frac{z^n}{n}.

The radius of convergence is 1.

We claim that the series converges for all z\in \mathbb{C} with |z|=1 and z\neq 1.

Theorem 3.44 Abel's test

Let (b_n)^\infty_{n=0} be a sequence such that:

(a) b_0\geq b_1\geq b_2\geq \cdots (non-increasing) (b) \lim_{n\to\infty}b_n=0

Then if |z|=1 and z\neq 1, \sum_{n=0}^\infty b_nz^n converges.

Proof:

Fix z\in \mathbb{C} with |z|=1 and z\neq 1.

Let a_n=z^n.

Then A_n=\sum_{k=0}^n z^k=\frac{1-z^{n+1}}{1-z}._

|A_n|\leq \frac{|1-z^{n+1}|}{|1-z|} for all n\in \mathbb{N}.

By triangle inequality, |1-z^{n+1}|\leq |1|+|z^{n+1}|=1+|z^{n+1}|.

And since |z|=1, |z^{n+1}|=|z|^{n+1}=1.

So |1-z^{n+1}|\leq 2.

So |A_n|\leq \frac{2}{|1-z|} for all n\in \mathbb{N}.

By Dirichlet's test, \sum_{n=0}^\infty b_nz^n

QED

Absolute convergence

The series \sum_{n=0}^\infty a_n is said to converge absolutely if \sum_{n=0}^\infty |a_n| converges.

If \sum_{n=0}^\infty a_n converges but does not converge absolutely, then \sum_{n=0}^\infty a_n is said to converge conditionally.

Absolute convergence are nice but conditionally convergent series are not.

Theorem 3.45 (Absolute convergence)

If \sum_{n=0}^\infty a_n converges absolutely, then \sum_{n=0}^\infty a_n converges.

Proof:

Use comparison test.

\sum_{n=0}^\infty |a_n|\geq \sum_{n=0}^\infty a_n

QED

Rearrangement of series:

Let f:\mathbb{N}\to \mathbb{N} be a bijection.

If \sum_{n=0}^\infty a_n is a sequence and b_n=a_{f(n)}, then (b_n)^\infty_{n=0} is a rearrangement of (a_n)^\infty_{n=0}.

If \sum_{n=0}^\infty a_n converges absolutely, then any rearrangement of \sum_{n=0}^\infty a_n converges to the same sum.

Example:

a_n=\frac{(-1)^{n+1}}{n}. b_n=a_{f(n)}.

n	1	2	3	4	5	6	7	8	9
f(n)	1	2	4	3	6	8	5	10	12
b_n	1	-1/2	-1/4	1/3	-1/6	-1/8	1/5	-1/10	-1/12

\sum_{n=1}^\infty a_n=\frac{1}{1}-\frac{1}{2}+\frac{1}{3}-\frac{1}{4}+\cdots=\log 2

\begin{aligned}
\sum_{n=1}^\infty b_n&=1-\frac{1}{2}-\frac{1}{4}+\frac{1}{3}-\frac{1}{6}-\frac{1}{8}+\frac{1}{5}-\frac{1}{10}-\frac{1}{12}+\cdots\\
&=\left(1-\frac{1}{2}\right)-\frac{1}{4}+\left(\frac{1}{3}-\frac{1}{6}\right)-\frac{1}{8}+\left(\frac{1}{5}-\frac{1}{10}\right)-\frac{1}{12}+\cdots\\
&=\frac{1}{2}-\frac{1}{4}+\frac{1}{6}-\frac{1}{8}+\frac{1}{10}-\frac{1}{12}+\cdots\\
&=\frac{1}{2}\left(1-\frac{1}{2}+\frac{1}{3}-\frac{1}{4}+\cdots\right)\\
&=\frac{1}{2}\log 2
\end{aligned}

You cannot always rearrange series.

But, if \sum_{n=0}^\infty a_n converges absolutely, then you can rearrange the series.

Theorem 3.55

Let (a_n)^\infty_{n=0} be a sequence in \mathbb{C} such that \sum_{n=0}^\infty |a_n| converges absolutely.

Then any rearrangement of \sum_{n=0}^\infty a_n converges absolutely to the same sum.

\sum_{n=0}^\infty a_n=\sum_{n=0}^\infty a_{f(n)}

Ideas of proof:

Let f:\mathbb{N}\to \mathbb{N} be a bijection.

and let b_n=a_{f(n)}.

Let s_n=\sum_{k=0}^n a_k,t_n=\sum_{k=0}^n b_k=\sum_{k=0}^n a_{f(k)}.

I_n=\{1,2,\cdots,n\}.

J_n=\{f(1),f(2),\cdots,f(n)\}.

\begin{aligned}
s_n-t_n&=\sum_{k=0}^n a_k-\sum_{k=0}^n a_{f(k)}\\
&=\sum_{k\in I_n} a_k-\sum_{k\in J_n} a_k\\
&= \sum_{k\in I_n\setminus J_n} a_k+\sum_{k\in J_n\setminus I_n} a_k\\
&\leq \sum_{k\in I_n\setminus J_n} |a_k|+\sum_{k\in J_n\setminus I_n} |a_k|
\end{aligned}

Key observation:

For every n\in \mathbb{N}, there exists a p such that \{1,2,\cdots,n\}\subset I_n\cap J_n.

Then |s_n-t_n|\leq \sum_{k=N+1}^\infty |a_k|.

QED