[Lecture Summary] 04 Real Anaysis : Series

This contents is based on Lecture of 18.100A of MIT Open Course. Coverage will be Lecture from 4~. link

Definition and Symbols

Series

• partial sum $s_m$. If $\limit s_m= s$ then we can treat it asa number
• Given sequence $x_n$, $\sum_{n=1}^{\infty} which is associated to $x_n$.
• If sequence $\lbrace s_m = \sum_{n=1}^{\infty} x_n\rbrace_{m=1}^{\infty}$ converges then series converges either!
• (RMK) Do not ahve to start seriest at $n=1$

Cauchy series

• sequence of partial sumis Cauchy

Absolutely convergence

• If $\sum \lvert x_n \rvert$ converges then $\sum x_n$ converges absolutely
• prove it via induction

Questions to be solved

1. Does $\limit x_n= 0$ implies $\sum x_n$ converges?

Theorem

If $\lvert r\rvert then \sum r^n$ converges and $\sum r^n= {1\over 1-r}$

Approach

• From definition via induction

Remark Series of the form from the above is called geometric series.

For sequence $x_n$ and $M\in\mathbb{N}$, $\sum_{n=1} x_n$ converges $\iff \sum_{n=M} x_n$ converges

$\sum x_n$ is Cauchy $\iff \sum x_n$ is convergent

Remark

• For sequence, the above relation is false.

$\sum x_n$ is Cauchy $\iff \forall \epsilon >0, \exists M\in\mathbb{N} s.t. \forall m\ge M and l>m, \lvert \sum_{n=m+1}^l x_n\rvert < \epsilon$

Remark

• Convergence(or Cauchu) of series depends on the tail of series

If $\sum x_n$$ converges then $\limit x_n=0$

$\sum {1\over n} does not converge

Approach

• Find the minimum of fraction of series.
• Then the minimum is unbounded!

Remark

• Series $\sum {1\over n} $ is called harmonic series

If $\forall n\in\mathbb{N}\ge 0$ then $\sum x_n$ convergence $\iff s_m$ is bounded.

$\sum x_n$ convergences absolutely then $\sum$ converges either.

Approach

• $\sum x_n$ is cauchy!

Remark

• $\sum {(-1)^n\over n}$ is convergent but not absolutely convergent!

(Comparison Test) $\forall n\in\mathbb{N}, 0\lex\ley$, If $\sum y_n$ converges then $\sum x_n$ converges!

(Comparison Test) $\forall n\in\mathbb{N}, 0\lex\ley$, If $\sum x_n$ duverges then $\sum y_n$ diverges!

For $p \in \mathbb{R}$, series $\sum {1\over n^p}$ converges $\iff p>1$

Approach

• Contradiction and Comparison Test!

(Ratio Test) Suppose $x_n \ne 0 \forall n$ and $L=\lim {\lvert x_{n+1}\over x_n}$ exist then if $L<1$,then $\sum x_n$ converges absolutely.

(Ratio Test) Suppose $x_n \ne 0 \forall n$ and $L=\lim {\lvert x_{n+1}\over x_n}$ exist then if $L>1$,then $\sum x_n$ diverges

Remark

• No inference for $L=1$

(Root test) Let $\sum x_n$ s.t. $L=\lim \lvert x_n\rvert^{1/n}. Then if $L<1$ then $\sum x_n$ converges absolutely

(Root test) Let $\sum x_n$ s.t. $L=\lim \lvert x_n\rvert^{1/n}. Then if $L>1$ then $\sum x_n$ diverges

(Alternating Sereis test) %x_n$ is a monotone decreasing sequence s.t. $x_n\to 0$. Then $\sum (-1)^n x_n$ converges