NoteNextra-origin/pages/Math4111/Math4111_L19.md

Lecture 19

Review

Binomial theorem: For n\in\mathbb{N},

(a+b)^n=\sum_{k=0}^{n}\binom{n}{k}a^{n-k}b^k

1. Show that 2^n\geq \binom{n}{4} for all n\geq 4. (Hint: Expand (1+1)^n using the binomial theorem) Proof:

   
   \begin{aligned}
   (1+1)^n&=\sum_{k=0}^{n}\binom{n}{k}1^{n-k}1^k\\
   &=\sum_{k=0}^{n}\binom{n}{k}\\
   &=\binom{n}{0}+\binom{n}{1}+\cdots+\binom{n}{n}\\
   &\geq\binom{n}{4}
   \end{aligned}
   

   EOP
2. Using part 1, show that \lim_{n\to\infty}\frac{n^3}{2^n}=0.
   Proof:

   
   \frac{n^3}{2^n}\leq\frac{n^3}{\binom{n}{4}}
   

   The value of \frac{n^3}{\binom{n}{4}} is decreasing when n\geq 4. EOP

New materials

Series

Definition 3.21

Let (a_n)_{n=1}^{\infty} be a sequence in \mathbb{C}. Let s_n=\sum_{k=1}^{n}a_k denotes the sequence of partial sums.

1. We say the series \sum_{n=1}^{\infty}a_n converges if the sequence of partial sums (s_n)_{n=1}^{\infty} converges.
2. We define the sum of the series \sum_{n=1}^{\infty}a_n to be the limit of the sequence of partial sums, i.e., \sum_{n=1}^{\infty}a_n=\lim_{n\to\infty}s_n=\lim_{n\to\infty}\sum_{k=1}^{n}a_k.

Theorem 3.22 (Cauchy criterion for series)

The series \sum_{n=1}^{\infty}a_n converges if and only if for every \epsilon>0, there exists N\in\mathbb{N} such that for all m,n\in\mathbb{N} with m\geq n\geq N,

\left|\sum_{k=n}^{m}a_k\right|<\epsilon.

Proof:

\sum_{n=1}^{\infty}a_n converges if and only if (s_n)_{n=1}^{\infty} converges.

Since \mathbb{C} is complete, (s_n)_{n=1}^{\infty} converges if and only if (s_n)_{n=1}^{\infty} is Cauchy.

Since (s_n)_{n=1}^{\infty} is Cauchy, for every \epsilon>0, there exists N\in\mathbb{N} such that for all m,n\in\mathbb{N} with m\geq n\geq N,

|s_m-s_n|=\left|\sum_{k=n}^{m}a_k\right|<\epsilon.

EOP

Special case of this theorem.

Corollary 3.23

If \sum_{n=1}^{\infty}a_n converges, then \lim_{n\to\infty}a_n=0.

Note: the converse is not true. Example: \sum_{n=1}^{\infty}\frac{1}{n} diverges.

The contrapositive of this corollary is: If \lim_{n\to\infty}a_n\neq 0, then \sum_{n=1}^{\infty}a_n diverges. It is useful naming as ``n-th term test for divergence''.

Observe:

\forall n,a_n\geq 0

(a_n) is a non-negative sequence if and only if (s_n)_{n=1}^{\infty} is increasing sequence.

So if (a_n) is a non-negative sequence, then \sum_{n=1}^{\infty}a_n converges if and only if (s_n)_{n=1}^{\infty} is bounded above.

Theorem 3.25 (Comparison test)

Let (a_n) be a sequence in \mathbb{C} and (c_n) be a non-negative sequence in \mathbb{R}. Suppose \forall n, |a_n|\leq c_n.

(a) If the series \sum_{n=1}^{\infty}c_n converges, then the series \sum_{n=1}^{\infty}a_n converges.
(b) If the series \sum_{n=1}^{\infty}a_n diverges, then the series \sum_{n=1}^{\infty}c_n diverges.

Proof:

(a) By Theorem 3.22, it's enough to show that for every \epsilon>0, there exists N\in\mathbb{N} such that for all m,n\in\mathbb{N} with m\geq n\geq N,

\left|\sum_{k=n}^{m}a_k\right|<\epsilon.

Let \epsilon>0 be arbitrary.

Since \sum_{n=1}^{\infty}c_n converges, by Theorem 3.22, for the above \epsilon, there exists N\in\mathbb{N} such that for all m,n\in\mathbb{N} with m\geq n\geq N,

\left|\sum_{k=n}^{m}c_k\right|\leq \sum_{k=n}^{m}c_k<\epsilon.

EOP

Theorem 3.26 (Geometric series)

Let x\in\mathbb{C}.

(a) If |x|<1, then the series \sum_{n=0}^{\infty}x^n converges and \sum_{n=0}^{\infty}x^n=\frac{1}{1-x}.
(b) If |x|\geq 1, then the series \sum_{n=0}^{\infty}x^n diverges.

Proof:

(b) If |x|\geq 1, then x^n does not converge to 0. So the series \sum_{n=0}^{\infty}x^n diverges.

(a) Let s_n=\sum_{k=0}^{n}x^k=1+x+x^2+\cdots+x^n.

xs_n=x+x^2+x^3+\cdots+x^n+x^{n+1}=s_n+x^{n+1}.

So s_n=\frac{1-x^{n+1}}{1-x}.

Since |x|<1, x^{n+1} converges to 0. So \lim_{n\to\infty}s_n=\frac{1}{1-x}.

EOP

Lemma 3.28

(a) \sum_{n=0}^{\infty}\frac{1}{n} diverges.
(b) \sum_{n=0}^{\infty}\frac{1}{n^2} converges.

Proof:

(a)

\begin{aligned}
\sum_{n=0}^{\infty}\frac{1}{n}&=\frac{1}{1}+\frac{1}{2}+\left(\frac{1}{3}+\frac{1}{4}\right)+\left(\frac{1}{5}+\frac{1}{6}+\frac{1}{7}+\frac{1}{8}\right)+\cdots\\
&>\frac{1}{2}+\frac{1}{2}+\left(\frac{1}{4}+\frac{1}{4}\right)+\left(\frac{1}{8}+\frac{1}{8}+\frac{1}{8}+\frac{1}{8}\right)+\cdots\\
&=\frac{1}{2}+\frac{1}{2}+\frac{1}{2}+\frac{1}{2}+\cdots\\
&=\infty
\end{aligned}

(b)

\begin{aligned}
\sum_{n=0}^{\infty}\frac{1}{n^2}&=\frac{1}{1}+\frac{1}{2^2}+\frac{1}{3^2}+\frac{1}{4^2}+\cdots\\
&<\frac{1}{1}+\left(\frac{1}{2^2}+\frac{1}{2^2}\right)+\left(\frac{1}{4^2}+\cdots+\frac{1}{4^2}\right)+\left(\frac{1}{8^2}+\cdots+\frac{1}{8^2}\right)+\cdots\\
&=\frac{1}{1}+\frac{2}{2^2}+\frac{4}{4^2}+\frac{8}{8^2}+\cdots\\
&=\frac{1}{1}+\frac{1}{2^2}+\frac{1}{2^3}+\frac{1}{2^4}+\cdots\\
&=\frac{1}{1-\frac{1}{2}}\\
&=\frac{1}{\frac{1}{2}}\\
&=2
\end{aligned}

Fun fact: \sum_{n=1}^{\infty}\frac{1}{n^2}=\frac{\pi^2}{6}.

EOP

Theorem 3.27 (Cauchy condensation test)

Suppose (a_n) is a non-negative sequence. The series \sum_{n=1}^{\infty}a_n converges if and only if the series \sum_{k=0}^{\infty}2^ka_{2^k} converges.

Proof:

Let s_n=\sum_{k=1}^{n}a_k and t_k=\sum_{k=0}^{k}2^ka_{2^k}.

If n\leq 2^k, then

\begin{aligned}
s_n&=a_1+a_2+\cdots+a_n\\
&\leq a_1+(a_2+a_3)+(a_4+a_5+\cdots+a_7)+\cdots+(a_{2^k}+a_{2^k+1}+\cdots+a_{2^{k+1}-1})\\
&\leq a_1+2a_2+4a_4+\cdots+2^ka_{2^k}\\
&=t_k.
\end{aligned}

If n\geq 2^{k+1}, then

\begin{aligned}
s_n&=a_1+a_2+\cdots+a_n\\
&\geq a_1+a_2+(a_3+a_4)+(a_5+a_6+\cdots+a_7)+\cdots+(a_{2^k}+a_{2^k+1}+\cdots+a_{2^{k+1}-1})\\
&\geq a_1+a_2+2a_4+\cdots+2^{k-1}a_{2^k}\\
&\geq \frac{1}{2}\left(a_1+2a_2+4a_4+\cdots+2^ka_{2^k}\right)\\
&=\frac{1}{2}t_k.
\end{aligned}

We have shown that

• If n\leq 2^k, then s_n\leq t_k.
• If n\geq 2^{k+1}, then s_n\geq \frac{1}{2}t_k.

So (s_n)_{n=1}^{\infty} is a bounded above.

By Theorem 3.14, (s_n)_{n=1}^{\infty} converges if and only if (t_k)_{k=0}^{\infty} converges.

EOP