MATH 3150 Class Notes 02-27-2025

notereal-analysis

Cauchy Sequence

Proposition 1

If $(x_n{)}_n$ convergent, then $(x_n{)}_n$ is Cauchy.

Proposition 2

If $(x_n{)}_n$ is Cauchy, then $(x_n{)}_n$ is convergent.

Proof

(Step 1) $(x_n{)}_n$ Cauchy $⟹$ $(x_n{)}_n$ bounded.

We know $\forall \epsilon >0$, $\exists N>0$ such that $\left|x_n-x_m\right|<\epsilon$, $\forall n,m>N$. From this we can fix $m>N$, and we get $x_m-\epsilon <x_n<x_m+\epsilon$. We can define $M_1=x_m+\epsilon$, and $M_2=x_m-\epsilon$, neither of which depend on $n$, which gives us $M_2<x_n<M_1$ for all $n>N$. Thus $x_n$ is bounded.

(Step 2) $(x_n{)}_n$ Cauchy $⟹$ $(x_n{)}_n$ convergent.

We know,

1. $\forall \epsilon >0$, $\exists N>0$ such that $\left|x_n-x_m\right|<\epsilon$ for any $n,m>N$.
2. $(x_n{)}_n$ is bounded (proved in step 1).

By the Bolzano-Weierstrass Theorem, from 2), we have $\exists (x_{n_k}{)}_k$ subsequence that converges to some limit $L$. Hence, $\forall \epsilon >0$, $\exists N_1>0$ such that $\left|x_{n_k}-L\right|<\epsilon$, $\forall k>N_1$.

We need to show $\left|x_n-L\right|<\epsilon$.

We can take $k>N_k$ large enough so that $n_k>N$, so we can take $m=n_k>N$ and get $\left|x_n-x_{n_k}\right|<\epsilon$ by 1). Thus,

$$\left|x_n-L\right|=\left|(x_n-x_{n_k})+(x_{n_k}-L)\right|\le \left|x_n-x_{n_k}\right|+\left|x_{n_k}-L\right|<\epsilon +\epsilon =2\epsilon ,$$

and we have that $(x_n{)}_n$ is convergent.

Applications to Series

Recall

We say ${\sum }_{n=1}^{\infty }{x}_n$ converges if the sequence $S_n=x_1+\cdots +x_n$ converges as $n\to \infty$.

For most series that are convergent, we are not able to find their sum.

When Do We Know a Series Converges

A series ${\sum }_{n=1}^{\infty }{x}_n$ converges,

1. If all $x_n\ge 0$ then ${\sum }_{n-1}^{\infty }{x}_n$ converges if and only if ${\sum }_{n=1}^{\infty }{x}_n<\infty$ (i.e. $S_n$ is bounded).
2. What if $x_n$s do not have a sign? If ${\sum }_{n=1}^{\infty }{x}_n$ is absolutely convergent, then ${\sum }_{n=1}^{\infty }{x}_n$ is convergent.