SAS重要证明结论

集合部分：

\(if \ A \subseteq B, \ B \subseteq A, \ then \ A = B\)
\((\cup_{\alpha \in A} S_{\alpha})^c = \cap_{\alpha \in A} (S_{a})^c\)

有序集部分：

\(if \ < is \ an \ order \ on \ S, \ these \ two \ followings \ hold:\)
\(1. \forall x,y \in S, \ x = y \ or \ x < y \ or \ y < x\)
\(2. if \ x < y, \ y < z, \ x < z\)

Suppose \((S, <)\) is an ordered set and has the least upper bound property. Then for any nonempty \(B \subseteq S\), which is bounded below, it has the greatest lower bound in \(S\).
Moreover, if \(L\) is the set of lower bounds of \(B\) in \(S\), then \(\sup L\) exists, and \(\sup L = \inf B\)

域部分：

Corollary:
\((F, <, +, \cdot)\) is an ordered field, then \(\forall 0 < x < y, n \in \mathbb{N}\), \(0 < x^n < y^n\) (proved by induction)

实数域与实数部分：

Archimedean property：
Suppose \(x,y \in R\) and \(x > 0\), there is some integer \(n\) such that \(nx > y\)
Denseness of \(\mathbb{Q}\) in \(\mathbb{R}\):
Suppose that \(x,y \in R\), and \(x < y\), there exists some rational numbers \(z\) that \(x < z < y\)
Corollary:
Suppose \(x,y \in R\) and \(\forall \epsilon \in \mathbb{Q}\) and \(\epsilon > 0\), if there are \(a_{\epsilon}, b_{\epsilon}\) such that \(x,y \in [a_{\epsilon}, b_{\epsilon}]\) and \(b_{\epsilon} - a_{\epsilon} < \epsilon\) , then \(x = y\).

序列

Limit definition:
If there exists an \(N\) such that \(\forall n > N,\epsilon > 0, l \in \mathbb{R}, | a_{n} - l | < \epsilon\), then we call \(l\) as the limit of the sequence.(\(a_{n} \to l\))

e.g. 1

Prove that if \(a_{n} = e^{b_{n}}\), \(\\lim_{ n \to \infty }b_{n} = C\), then \(\\lim_{ n \to \infty }a_{n} = e^C\)
we see that \(\\lim_{ n \to \infty }b_{n} = C\)
therefore, there exists an \(N\) such that \(\forall n \geq N, \epsilon > 0, |b_{n}-C| < \epsilon\)
therefore, \(\forall n \geq n\),
\(|a_{n} - e^C| = |e^{b_{n}} - e^C| = e^{C}|(e^{b_{n} - C} - 1)|\)
because \(\forall \epsilon > 0, |b_{n} - C| < \epsilon\)
we set \(\epsilon' = \ln(1 + \frac{\epsilon}{e^C})\)
when \(|b_{n} - C| \leq \epsilon'\)
\(e^{|b_{n} - C|} \leq e^{\epsilon'} = 1 + \frac{\epsilon}{e^C}\)
\(e^C|e^{b_{n} - C} - 1| \leq e^C(1 + \frac{\epsilon}{e^C} - 1) = \epsilon\)
therefore we get \(\lim_{ n \to \infty }a_{n} = e^C\)

e.g. 2

Let \(x_{n} = \sqrt[n]{n}\)
Find the limit of \((x_{n})\)
We see that \(x_{n} = e^{\frac{\ln n}{n}}\)
through example shown above, we only need to prove that \(\\lim_{ n \to \infty} \frac{\ln n}{n} = 0\)
we set \(f(x) = \sqrt{ n } - \ln n\)
\(f'(x) = \frac{1}{2\sqrt{ x }} - \frac{1}{x} = \frac{\sqrt{ x } - 2}{2x}\)
when \(x > 4\), \(f'(x) > 0\)
while \(f(4) = 2 - \ln 4 = 2 - 2\ln 2 = 2(1 - \ln 2) > 0\)
therefore we get when \(x > 4\), \(f(x) > 0\)
thus \(\ln n < \sqrt{ n }\)
when \(x \to \infty\), \(\ln n < \sqrt{ n }\)
\(0 < \frac{\ln n}{n} < \frac{\sqrt{ n }}{n}\)
Because \(\\lim_{ n \to \infty } \frac{1}{\sqrt{ n }} = 0, \lim_{ n \to \infty } 0 = 0\)
thus \(\\lim_{ n \to \infty } \frac{\ln n}{n} = 0\)
Combined with e.g.1, we get \(\\lim_{ n \to \infty } \sqrt[n]{ n } = \\lim_{ n \to \infty }e^{\frac{\ln n}{n}} = e^{0} = 1\)

序列与单调性

Monotone Convergence Theorem
Suppose that \(\{a_{n}\}\) is monotonically sequence. Then, \(\{a_{n}\}\) is convergent if and only if it is bounded.
By reduction to absurdity we can easily prove it.
e.g.
Show that if \(s \in \mathbb{R}\) and \(|\lambda| > 1\), \(\lim_{ n \to \infty }a_n =\frac{n^s}{\lambda^{n}} = 0\)
Pf.
because we need to prove that \(|\frac{n^s}{\lambda^{n}} - 0| = \frac{n^s}{|\lambda|^{n}} \to 0\)
so we only have to focus on when \(\lambda > 0\), and we assume that in the followings:
Firstly we get \(\frac{a_{n+1}}{a_{n}} = \frac{(n+1)^s}{\lambda^{n+1}} \times \frac{\lambda^{n}}{n^s} = \frac{1}{\lambda} \times (\frac{n+1}{n})^s\)
we let \(RHS < 1\), then we have \(\frac{n+1}{n} < \sqrt[s]{ \lambda }\)
\(n + 1 < \sqrt[s]{ \lambda }n\)
when \(n \geq \lceil \frac{1}{\sqrt[s]{ \lambda } - 1} \rceil\),
\(\{a_{n}\}\) is a monotonically decreasing sequence.
The next step is to prove that it is bounded.
It is trivial that \(a_{n} > 0\) and its upper bound is \(a_{\lceil \frac{1}{\sqrt[s]{ \lambda } - 1} \rceil}\)
so it's bounded.
Q.E.D
Bernoulli Inequality
\(\forall x \geq -1, n \in \mathbb{N}, (1+x)^n \geq 1 + nx\)
Pf.
By induction.
Firstly, we see that when \(n = 1\), \(LHS = RHS\), so this inequality holds.
when \(x \geq 0\), by binomial theorem, \(LHS = \sum_{i = 0}^{n}{\binom{n}{i}x^i} = 1 + nx + \sum_{i=2}^{n}{\binom{n}{i}x^i} \geq 1 + nx\)
it suffices to show that when \(x \in [-1,0)\), this inequality still holds
let's consider when \(n' = n + 1\)
we set \(t = 1 + x\), then \(t \in [0, 1)\), we times \(t\) in both sides, then we get
\((1+x)^{n+1} \geq (1+x)(1+nx)=1+nx+x+nx^2 \geq 1+(n+1)x\)
So it holds for \(n'=n+1\)
Q.E.D

自然底数\(e\)

小结论。
\(\{a_{n} = (1 + \frac{1}{n})^n\}\) is monotonically increasing while \(\{b_{n} = (1 + \frac{1}{n})^{n+1}\}\) is monotonically decreasing, and also \((1 + \frac{1}{n})^n < e < (1+\frac{1}{n})^{n+1}\) because these two sequence converge to the same limit \(e\)
and then we could get \(n\ln(1+\frac{1}{n}) < 1 < (n+1)\ln(1+\frac{1}{n})\)
thus \(\frac{1}{n+1} < \ln(1+\frac{1}{n}) < \frac{1}{n}\)

Stolz 定理

Stolz Theorem
Suppose \((y_{n})\) is strictly increasing and diverges to \(\infty\). Let \((x_{n})\) be another sequence in \(\mathbb{R}\). Suppose \(\lim_{ n \to \infty }\frac{x_{n+1}-x_{n}}{y_{n+1}-y_{n}} = l\), then \(\lim_{ n \to \infty }\frac{x_{n}}{y_{n}} = l\)
It's just like a discrete version of L'hopital rule.
Pf.
Let \(\epsilon > 0\) be fixed, \(\exists N\) such that \(\forall n \geq N,l - \frac{\epsilon}{3} < \frac{x_{n+1}-x_{n}}{y_{n+1}-y_{n}} < l + \frac{\epsilon}{3}\)
then \((l-\frac{\epsilon}{3})(y_{n+1}-y_{n}) < x_{n+1} - x_{n} < (l + \frac{\epsilon}{3})(y_{n+1}-y_{n})\)
and \(\left( l-\frac{\epsilon}{3} \right)\sum_{i=N}^{n}(y_{i+1}-y_{i}) < \sum_{i=N}^{n}(x_{{i+1}}-x_{i}) < \left( l+\frac{\epsilon}{3} \right)\sum_{i=N}^{n}(y_{i+1}-y_{i})\)
thus \((l-\frac{\epsilon}{3})(y_{n+1} - y_{N}) < x_{n+1}-x_{N} < (l+\frac{\epsilon}{3})(y_{n+1} - y_{N})\)
and \((l-\frac{\epsilon}{3})(y_{n+1} - y_{N}) + x_{N}< x_{n+1} < x_{N} + (l+\frac{\epsilon}{3})(y_{n+1} - y_{N})\)
we divide this inequality by \(y_{n+1}\)
thus \((l-\frac{\epsilon}{3})(1 - \frac{y_{N}}{y_{n+1}}) + \frac{x_{N}}{y_{n+1}}< \frac{x_{n+1}}{y_{n+1}} < \frac{x_{N}}{y_{n+1}} + (l+\frac{\epsilon}{3})(1 - \frac{y_{N}}{y_{n+1}})\)
because \(\lim_{ n \to \infty } y_{n} = \infty\), so any fraction with \(y_{n}\) in the bottom diverges to \(0\)
thus \(\exists N_{1}\) such that \(\forall n \geq N_{1}, \frac{x_{N}}{y_{n+1}} < \frac{\epsilon}{3}\)
thus \(\frac{x_{n+1}}{y_{n+1}} < \frac{x_{N}}{y_{n+1}} + (l+\frac{\epsilon}{3})(1 - \frac{y_{N}}{y_{n+1}}) < \frac{x_{N}}{y_{n+1}} + (l+\frac{\epsilon}{3})(1 - 0) < \frac{\epsilon}{3} + l + \frac{\epsilon}{3} < l + \epsilon\)
similarly, \(\exists N_{2}\) such that \(\forall n \geq N_{2}, (l+\frac{\epsilon}{3})\frac{y_{N}}{y_{n+1}} < \frac{\epsilon}{3}\)
thus \(\frac{x_{n+1}}{y_{n+1}} > (l-\frac{\epsilon}{3})(1 - \frac{y_{N}}{y_{n+1}}) + \frac{x_{N}}{y_{n+1}} > (l-\frac{\epsilon}{3})(1 - \frac{y_{N}}{y_{n+1}}) = l-\frac{\epsilon}{3} - (l+\frac{\epsilon}{3})\frac{y_{N}}{y_{n+1}} > l-\frac{\epsilon}{3} - \frac{\epsilon}{3} > l+\epsilon\)
We simultaneously proved these two inequalities with different directions by amplifying and minifying the positive one or the negative one to 0.
then we set \(n \geq \max\{N_{1},N_{2}\}\),
\(l-\epsilon < \frac{x_{n+1}}{y_{n+1}}<l+\epsilon\)
thus we get \(\lim_{ n \to \infty }\frac{x_{n}}{y_{n}} = l\)

子序列

Subsequence is defined to choose some of the element from the sequence without changing their original order.
Specifically, if \(\{a_{n}\}\) is a subsequence of \(\{b_{n}\}\), we can have \(a_{n} = b_{c_{n}}\), and \(\{c_{n}\}\) is strictly monotonically increasing.
There's a useful theorem, by using which we can prove a sequence doesn't converge.
Theorem:
If \(\lim_{ n \to \infty }a_{n} = l\), then every subsequence of \(\{a_{n}\}\) converges to \(l\)
Therefore if we have two subsequences with different limits, then the original sequence does not converge.

柯西收敛准则

Theorem:
For a sequence of real numbers, it must have a monotone subsequence.
And this theorem lead to a stronger theorem.
Bolzano-Weierstrass Theorem:
Every bounded sequence has a convergent subsequence.
Bolzano-Cauchy convergence criterion:
Let \(\{a_{n}\}\) be arbitrarily fixed, it converges if and only if there exists an integer \(N\) for all \(\epsilon > 0\) such that \(\forall n,m \geq N, |a_{n}-a_{m}| < \epsilon\).
Theorem:\(\lim_{ n \to \infty } \sup a_{n}=\lim_{ n \to \infty }\inf a_{n} \Longleftrightarrow \lim_{ n \to \infty } a_{n} =l\)

级数

Actually it's just a prefix sum of an array or a sequence.
\(\sum_{k=1}^{\infty}{a_{k}}\) is called a series.
if \(\{s_{n}\}\) converges then \(\lim_{ n \to \infty }a_{n} = 0\).
\(\lim_{ n \to \infty }a_{n} = \lim_{ n \to \infty }(s_{n}-s_{n-1})=\lim_{ n \to \infty}s_{n} - \lim_{ n \to \infty }s_{n-1}= C -C =0\)
Comparison Test
if \(0 < a_{n} < b_{n}\),then if \(\sum a_{n}\) diverges, so does \(\sum b_{n}\)
if\(\sum b_{n}\) converges, so does \(\sum a_{n}\)
Comparison Test'
if \(\lim_{ n \to \infty }\frac{a_{n}}{b_{n}} =l\), then \(\sum a_{n} \ converges \Longleftrightarrow \sum b_{n} \ converges\)

e.g. 3

A little conclusion
\(\sum_{n=1}^{\infty}r^{n-1}\) converges if \(|r| < 1\)
\(\sum r^{n} = \frac{1-r^{n+1}}{1-r}\)
\(\lim_{ n \to \infty } \frac{1-r^{n+1}}{1-r} = \frac{1}{1-r}\) if \(|r| < 1\)
Corollary
\(\sum_{n=1}^{\infty} \frac{1}{n^p}\) converges if \(p > 1\),otherwise it diverges.
Pf.
It is obvious that if \(p < 1\),\(\frac{1}{n^p} > \frac{1}{n}\)
and \(\{a_{n} = \frac{1}{n}\}\) (harmonic series) is clearly divergent.
This can be proved by minifying. Proof will be shown below.
By comparison test it's clear that \(\sum_{n=1}^{\infty} \frac{1}{n^p}\) diverges when \(p < 1\)
when it comes to the situation when \(p > 1\)
we can divide the sum into different parts.
we divide them into groups
we order that \(S_{1}=\{2\},S_{2}=\{3,4\},S_{3}=\{5,6,7,8\},S_{n}=\{2^{n-1}+1,2^{n-1}+2,\dots,2^n\}\)
in one group \(\frac{1}{n^p} > \frac{1}{m^p}\) if \(m > n\)
so \(\forall m \in S_{n}, \frac{1}{(2^{n-1})^p} \geq \frac{1}{m^p}\)
we sum up all element in one single group:
\(\sum_{k=2^{n-1}+1}^{2^n}\frac{1}{k^p} < \sum_{k=2^{n-1}+1}^{2^n}\frac{1}{({2^{n-1}})^p}\)
\(\sum_{k=1}^{2^m}\frac{1}{k^p}=1+\sum_{n=1}^{m}\sum_{i=2^{n-1}+1}^{2^n}\frac{1}{i^p} < 1+\sum_{n=1}^{m}\sum_{i=2^{n-1}+1}^{2^n}\frac{1}{({2^{n-1}})^p} = 1+ \sum_{n=1}^m 2^{n-1} \frac{1}{2^{np-p}}= 1+\sum_{n=1}^m \frac{1}{2^{p}}\)
therefore what it is smaller than is totally a geometric sequence thus it converges.
Q.E.D

Harmonic sequence convergence

Clearly through inequality \(\ln(1+\frac{1}{n}) < \frac{1}{n}\) we see that
\(\sum_{n=1}^{m}\frac{1}{n} > \sum_{n=1}^{m}\ln(1+\frac{1}{n})=\sum_{n=1}^{m}(\ln(1+n)-\ln n)=\ln(m+1)-\ln 1=\ln(m+1)\)
and \(\{a_{n}=\ln (n+1)\}\) diverges thus \(\sum_{n=1}^{\infty}\frac{1}{n}\) diverges.

e.g. 4

Like what we've learnt in senior high.
\(\sum_{n=1}^{\infty}\frac{1}{n(n+n_{0})}\) converges
Pf.
\(\frac{1}{n(n+n_{0})}=\frac{1}{n_{0}}(\frac{1}{n}-\frac{1}{n+n_{0}})\)
\(\sum_{n=1}^{m}\frac{1}{n(n+n_{0})}=\sum_{n=1}^{m}\frac{1}{n_{0}}\left( \frac{1}{n}-\frac{1}{n+n_{0}} \right)=\frac{1}{n_{0}}\left( \sum_{n=1}^{n_{0}}\frac{1}{n}-\sum_{n=m+1}^{m+n_{0}}\frac{1}{n} \right)\)
\(\sum_{n=1}^{\infty}\frac{1}{n(n+n_{0})}=\lim_{ m \to \infty }\frac{1}{n_{0}}\left( \sum_{n=1}^{n_{0}}\frac{1}{n}-\sum_{n=m-n_{0}}^{m}\frac{1}{n} \right) = \frac{1}{n_{0}}\sum_{n=1}^{n_{0}}\frac{1}{n}\)

绝对收敛

A series is called to be absolutely convergent if \(\sum_{n=1}^{\infty}|a_{n}|\) converges, and \(\sum_{n=1}^{\infty}a_{n}\) converges.
A series is called to be conditionally convergent if \(\sum_{n=1}^{\infty}|a_{n}|\) diverges and \(\sum_{n=1}^{\infty}a_{n}\) converges.
Theorem:
If \(\sum_{n=1}^{\infty}|a_{n}|\) converges, so does \(\sum_{n=1}^{\infty}a_{n}\).
Pf.
Because we don't know what its limit is, it give us a hint to use Cauchy criterion
\(\forall \epsilon >0, \exists N, \forall n,m \geq N,|\sum_{i=1}^n|a_{i}|-\sum_{i=1}^m|a_{i}||<\epsilon\)
and \(|\sum_{i=1}^na_{i}-\sum_{i=1}^ma_{i}|=|\sum_{i=m+1}^na_{i}| \leq \sum_{i=m+1}^n|a_{i}|=|\sum_{i=m+1}^n|a_{i}||=|\sum_{i=1}^n|a_{i}|-\sum_{i=1}^m|a_{i}||<\epsilon\)

级数和重排

Theorem(Dirichlet):
If \(\{b_{n}\}\) is a rearrangement of \(\{a_{n}\}\),and \(\sum_{n=1}^{\infty}a_{n}\) is absolutely convergent,then the following statements hold:

1. \(\sum_{n=1}^{\infty}b_{n}\) is absolutely convergent
2. \(\sum_{n=1}^{\infty}a_{n}=\sum_{n=1}^{\infty}b_{n}\)
   Theorem(Riemann):
   If \(\sum_{n=1}^{\infty}a_{n}\) is conditionally convergent and \(\forall \gamma \in [-\infty,\infty]\), there exist a \(b_{n}\) such that \(\sum_{n=1}^{\infty}b_{n}=\gamma\) where \(b_{n}=a_{\sigma(n)}\)

级数之积（卷积）

Theorem(Cauchy):
Given a bijection \(s \to (n_{s},k_{s})\) from \(\mathbb{N}\) onto \(\mathbb{N} \times \mathbb{N}\)
\(\sum_{s=1}^{\infty}a_{n_{s}}b_{k_{s}}=\left( \sum_{n=1}^{\infty}a_{n} \right)\left( \sum_{n=1}^{\infty}b_{n} \right)\)

敛散性检测

d’Alembert test and ratio test:
Consider a strictly positive series.
(a) if \(\exists l \in (0,1)\),\(\forall i,\frac{a_{i+1}}{a_{i}} \leq l\),then the series converges.
(b) if \(\forall i,\frac{a_{i+1}}{a_{i}} \geq 1\),the the series diverges.
More specifically, there's ratio test:
For a strictly positive series, if \(\lim_{ n \to \infty }\frac{a_{n+1}}{a_{n}}=l\)
if\(l<1\),the series converges, if\(l > 1\), the series diverges.
Cauchy test and root test:
(a) If for some natural numbers \(N\) such that \(\forall n \geq N,(a_{n})^{\frac{1}{n}} \geq l <1\), then the series converges.
(b) If for some natural numbers \(N\) such that \(\forall n \geq N,(a_{n})^{\frac{1}{n}} \geq 1\), then the series diverges.
Similarly to ratio test above, root test is described as follows:
Consider \(\lim_{ n \to \infty }(a_{n})^{\frac{1}{n}}=l\)
if \(l<1\), the series converges, if \(l>1\) the series diverges.
Theorem(Leibniz):
\(\sum_{n=1}^{\infty}(-1)^na_{n}\) converges if \(\{a_{n}\}\) is nonincreasing and converges to \(0\).
A theorem of Cauchy:
For a nonincreasing sequence \(\{a_{n}\}\),\(\lim_{ n \to \infty }a_{n}=0\),\(\sum_{n=1}^{\infty}a_{n}\)diverges \(\Longleftrightarrow\) \(\sum_{k=1}^{\infty}2^ka_{2^k}\) converges.

幂级数

We define \(\sum_{n=1}^{\infty}a_{n}x^n\) as a power series.