微积分(一)笔记1

前言：这是暑假在B站温习微积分找到的教程，自己做了一些笔记（有时间就翻译）和习题（没有答案，自己做的不保证对），现记录如下。
神秘学校 微积分一 (The School You Don’t Know Which, CASE)

实数理论

omit…

数列极限的定义与性质

Def  Let $a_n(n\in N)$ be a sequence in $R$ and $L\in R$, we say $a_n$ converge to $L(asn\to \infty )$ if
$$\forall ϵ>0\exists N_1\in Ns.t.\left[n\ge N_1⟹|a_n-L|<ϵ\right]$$
Terminology: If such $L$ exists (doesn’t exist), we call it the limit of $a_n$ and we call $a_n(n\in N)$ a convergent(divergent) sequence.
Notation:
$$\underset{n\to \infty }{\lim }{a}_n=L$$
Some generalized notations:
$$\underset{n\to \infty }{\lim }{a}_n=\infty (-\infty )\stackrel{Def}{⟺}\forall M>0\exists N_1\in Ns.t.\left[n\ge N_1⟹a_n\ge M(a_n\le M)\right]$$
Ex.
(1) Show that if $\underset{n\to \infty }{\lim }{a}_n=L$ and $\underset{n\to \infty }{\lim }{a}_n=M$ then $L=M$.
Proof: Suppose $L\ne M$ and $L>M$ and let $d=L-M>0$.
By definition we have
$\forall ϵ>0\exists N_1,N_2\in N[n\ge N_1⟹|a_n-L|<ϵ$ and $n\ge N_2⟹|a_n-M|<ϵ]$
Let N 3 = max ⁡ { N 1 , N 2 } , n ≥ N 3    ⟹    N_3 = \max\left\{N_1,N_2 \right\}, n\geq N_3 \implies N3​=max{N1​,N2​},n≥N3​⟹
$$\begin{gathered} \{\begin{array}{r}|a_n-L|<ϵ\\ |a_n-M|<ϵ\end{array}holds⟹|a_n-L-a_n+M|\le |a_n-L|+|a_n-M|<2ϵ<d \\ ⟹|M-L|<d=L-M\ast \end{gathered}$$
This leads to a contradiction, so original proposition is proved.#


(2) Show $a_n(n\in N^{\ast })$ is convergent    ⟹    { a n ∣ n ∈ N ∗ } \implies \left\{a_n|n\in N^*\right\} ⟹{an​∣n∈N∗} is bounded.
Proof: Suppose $a_n$ converges to $L$.
$$\forall ϵ>0\exists N\in N^{\ast }s.t.n\ge N⟹|a_n-L|<ϵ⟹L-ϵ<a_n<L+ϵ$$
Let M 1 = max ⁡ { ∣ L − ϵ ∣ , ∣ L + ϵ ∣ }    ⟹    ∣ a n ∣ < M 1 ( n ≥ N ) M_1 = \max\{|L-\epsilon|,|L+\epsilon|\}\implies |a_n|<M_1(n\geq N) M1​=max{∣L−ϵ∣,∣L+ϵ∣}⟹∣an​∣<M1​(n≥N),and M 2 = max ⁡ n < N { a n } , M = max ⁡ { M 1 , M 2 }    ⟹    ∣ a n ∣ < M M_2 =\max\limits_{n<N}\{a_n\},M = \max\{M_1,M_2\}\implies |a_n|<M M2​=n<Nmax​{an​},M=max{M1​,M2​}⟹∣an​∣<M #


(3) Show that if $a_n\le b_n$ for all $n\in N^{\ast }$ and $\underset{n\to \infty }{\lim }{a}_n=L,\underset{n\to \infty }{\lim }{b}_n=M$,then $L\le M$.
(Q: What if “$\le$” is replaced by “$<$”? A: The property doesn’t hold. Consider $a_n=\frac{1}{n^2+2}$ and $b_n=\frac{1}{n^2+1}$)
Proof. Suppose $L>M$. We have
∀ ϵ > 0    ∃ N 1 , N 2 ∈ N ∗    s . t .    n > max ⁡ { N 1 , N 2 }    ⟹    { ∣ a n − L ∣ < ϵ ∣ b n − M ∣ < ϵ    ⟹    L − M = ∣ M − L ∣ < ∣ a n − b n + M − L ∣ = ∣ ( a n − L ) − ( b n − M ) ∣ < ∣ a n − L ∣ + ∣ b n − M ∣ < 2 ϵ \forall \epsilon>0\;\exists N_1,N_2\in N^* \;s.t.\;n>\max\{N_1,N_2\}\implies \begin{cases}|a_n-L|<\epsilon\\|b_n-M|<\epsilon\end{cases}\implies L-M=|M-L|<|a_n-b_n+M-L|=|(a_n-L)-(b_n-M)|<|a_n-L|+|b_n-M|<2\epsilon ∀ϵ>0∃N1​,N2​∈N∗s.t.n>max{N1​,N2​}⟹{∣an​−L∣<ϵ∣bn​−M∣<ϵ​⟹L−M=∣M−L∣<∣an​−bn​+M−L∣=∣(an​−L)−(bn​−M)∣<∣an​−L∣+∣bn​−M∣<2ϵ
Let $ϵ=\frac{L-M}{3}⟹\frac{2(L-M)}{3}>L-M\ast$
This leads to a contradiction, original property holds.


Remark: Changing or removing finitely many terms in $a_n$ doesn’t effct $a_n(n\in N^{\ast })$'s being convergent (and it’s limit) / divergent.

Propsition:If $\underset{n\to \infty }{\lim }{a}_n=L$ and $\underset{n\to \infty }{\lim }{b}_n=M$ then
(1) $\underset{n\to \infty }{\lim }(a_n\pm b_n)=L\pm M$;
(2) $\underset{n\to \infty }{\lim }(a_n{b}_n)=LM$
(3) If $M\ne 0$,then $b_n\ne 0$ for all but finitely many n, and $\underset{n\to \infty }{\lim }\frac{a_n}{b_n}=\frac{L}{M}$
（如果$M\ne 0$ 那么只有有限多项 $b_n=0$，并且$\underset{n\to \infty }{\lim }\frac{a_n}{b_n}=\frac{L}{M}$）


Example. If $a>1$, then $\underset{n\to \infty }{\lim }\frac{1}{a^n}=0$
$\frac{1}{a^n}=\frac{1}{(1+(a-1){)}^n}\le \frac{1}{1+nb}$


Ex.
If $a_n\le c_n\le b_n(n\in N^{\ast })$,$\underset{n\to \infty }{\lim }{a}_n=L$ and $\underset{n\to \infty }{\lim }{b}_n=L$, then $\underset{n\to \infty }{\lim }{c}_n=L$
Proof. $$\begin{gathered} \forall ϵ>0\exists N\in N^{\ast }s.t.n\ge N⟹\{\begin{array}{l}|a_n-L|<ϵ\\ |b_n-L|<ϵ\end{array}⟹\{\begin{array}{l}{c}_n\le b_n<L+ϵ\\ c_n>a_n>L-ϵ\end{array} \\ ⟹|c_n-L|<ϵ⟹\underset{n\to \infty }{\lim }{c}_n=L \end{gathered}$$  #

单调数列的收敛性

Def  A seq. $a_n(n\in N^{\ast })$ in $R$ is
(1) nondecreasingly monotone(单调非减) / increasing if $a_n\le a_{n+1}$ for all $n\in N^{\ast }$, denoted $a_n\nearrow$.
(resp. nonincreasingly monotone(单调非增) / decreasingly if $a_n\ge a_{n+1}$ for all $n\in N^{\ast }$, denoted $a_n\searrow$)
(2) strictly increasingly (resp. strictly decreasing) if
$$\forall n\in N^{\ast }[a_n<(resp.>)a_{n+1}]$$
denoted $a_n\nearrow \nearrow ($$a_n\searrow \searrow$).
Theorem. Boundness from above + $\nearrow$ $⟹$ convergence