Advanced Math Math Analysis |01 Limits, Continuous

Advanced Math & Math Analysis |01 Limits, Continuous

Note: For simple definitions and corollaries, we will not elaborate.

Some easy knowledge

Mathematical induction

Theorem
Let $T(n)$ is a proposition if:

(1) $T(1)$ is true;

(2)let $T(k)(k\ge 1)$ is true;

(3) prove: $T(k+1)$is true.

So $T(n)$ is true.

Example

Let $n\ge 1$ and $n\in \mathbb{N}$,prove

$$n!<{\left(\frac{n+1}{2}\right)}^n$$

Proof

When $n=2$, $2!=2<\frac{9}{4}={\left(\frac{2+1}{2}\right)}^2$, and the proposition holds.

Suppose the proposition holds when $n=k(k\ge 2)$, that is, $$k!<{\left(\frac{k+1}{2}\right)}^k$$

When $n=k+1$,

$$\begin{array}{rl}(k+1)!& =k!(k+1)\\ & <{\left(\frac{k+2}{2}\cdot \frac{k+1}{k+2}\right)}^k\cdot (k+1)\\ & ={\left(\frac{k+2}{2}\right)}^k\cdot \frac{(k+1{)}^{k+1}}{(k+2{)}^k}\\ & =\frac{(k+2{)}^{k+1}}{2^k}\cdot {\left(\frac{k+1}{k+2}\right)}^{k+1}\\ & =2{\left(\frac{k+2}{2}\right)}^{k+1}\cdot \frac{1}{{\left(\frac{k+2}{k+1}\right)}^{k+1}}\\ & =2{\left(\frac{k+2}{2}\right)}^{k+1}\cdot \frac{1}{{\left(1+\frac{1}{k+1}\right)}^{k+1}}\end{array}$$

By $(1+x{)}^n\ge 1+nx$,$x\in (-1,+\infty )$,$n\in {\mathbb{N}}_{+}$, we have

$${\left(1+\frac{1}{k+1}\right)}^{k+1}>1+(k+1)\cdot \frac{1}{k+1}=2$$ So $$(k+1)!<2{\left(\frac{k+2}{2}\right)}^{k+1}\cdot \frac{1}{2}={\left(\frac{k+2}{2}\right)}^{k+1}$$

Thus, when $n=k+1$, the proposition holds. The proof is completed.

Some equation of trigonometric function

Inequality

Limits of sequence

Example Prove an open interval has no maximum value.

Proof

For any open continue interval $(a,b)$ can be mapping to $(0,1)$ , by

$$f:\frac{x-a}{b-a}$$

Suppose there exists a $\max {(0,1)} \stackrel{\scriptstyle\triangle}{=} \alpha $ ,we have

$$\begin{array}{rl}& 0<\frac{1}{2}\alpha <\frac{1}{2}\\ & \Rightarrow \frac{1}{2}<\frac{\alpha +1}{2}<1\end{array}$$

Obviously,

$$\alpha <\frac{\alpha +1}{2}$$

This contradicts the assumption, so the assumption is wrong.

Example Let T={x∣x∈Qand x>0,x2<2}T = \{ x \mid x \in \mathbb{Q} \text{ and } x > 0, x^2 < 2 \}T={x∣x∈Q and x>0,x2<2}. Prove that $T$ has no upper bound within $\mathbb{Q}$.

Proof Use proof by contradiction. Suppose $T$ has a least upper bound within $\mathbb{Q}$. Denote $\sup T=\frac{n}{m}$ (where $m,n\in {\mathbb{N}}^{\ast }$ and $m,n$ are coprime). Then, it is obvious that:

$$1<{\left(\frac{n}{m}\right)}^2<3$$

Since the square of a rational number cannot equal 2, there are only the following two possibilities: (1) $1<{\left(\frac{n}{m}\right)}^2<2$: Let $2-\frac{n^2}{m^2}=t$, then $0<t<1$. Let $r=\frac{n}{6m^2}t$. Obviously, $\frac{n}{m}+r>0$, and $\frac{n}{m}+r\in \mathbb{Q}$. Since $r^2=\frac{n^2}{36m^4}{t}^2<\frac{1}{18}t$, and $\frac{2n}{m}r=\frac{n^2}{3m^3}t<\frac{2}{3}t$, we can obtain:
$${\left(\frac{n}{m}+r\right)}^2-2=r^2+\frac{2n}{m}r-t<0$$
This shows that $\frac{n}{m}+r\in T$, which contradicts the fact that $\frac{n}{m}$ is the least upper bound of $T$. (2) $2<{\left(\frac{n}{m}\right)}^2<3$: Let $\frac{n^2}{m^2}-2=t$, then $0<t<1$. Let $r=\frac{n}{6m^2}t$. Obviously, $\frac{n}{m}-r>0$, and $\frac{n}{m}-r\in \mathbb{Q}$. Since $\frac{2n}{m}r=\frac{n^2}{3m^3}t<t$, we can obtain:
$${\left(\frac{n}{m}-r\right)}^2-2=r^2-\frac{2n}{m}r+t>0$$
This shows that $\frac{n}{m}-r$ is also an upper bound of $T$, which contradicts the fact that $\frac{n}{m}$ is the least upper bound of $T$.

Example Caculate

$$\underset{n\to \infty }{\lim }\frac{(2n-1)!!}{(2n)!!}$$
and
$$\underset{n\to \infty }{\lim }\sqrt{n}\frac{(2n-1)!!}{(2n)!!}$$

Solve

(1)

$$\boxed{\underset{n\to \infty }{\lim }\frac{(2n-1)!!}{(2n)!!}=0}$$

$$\begin{array}{rl}0\le \underset{n\to \infty }{\lim }\frac{(2n-1)!!}{(2n)!!}& =\underset{n\to \infty }{\lim }\frac{{\prod }_{i=1}^n(2i-1)}{{\prod }_{i=1}^n(2i)}\\ & =\sqrt{\underset{n\to \infty }{\lim }{\left(\frac{{\prod }_{i=1}^n(2i-1)}{{\prod }_{i=1}^n(2i)}\right)}^2}\\ & \le \sqrt{\underset{n\to \infty }{\lim }\frac{{\left({\prod }_{i=1}^n(2i-1)\right)}^2}{{\prod }_{i=1}^n(2i+1){\prod }_{i=1}^n(2i-1)}}\\ & =\sqrt{\underset{n\to \infty }{\lim }\frac{{\prod }_{i=1}^n(2i-1)}{{\prod }_{i=1}^n(2i+1)}}\\ & =\sqrt{\underset{n\to \infty }{\lim }\frac{1}{2n+1}}\\ & =0\end{array}$$

(2)

$$\boxed{\underset{n\to \infty }{\lim }\sqrt{n}\frac{(2n-1)!!}{(2n)!!}=\frac{\sqrt{\pi }}{2}}$$

$$\begin{array}{rl}& {\int }_0^{\frac{\pi }{2}}{\sin }^{2n}x\mathrm{d}x{\int }_0^{\frac{\pi }{2}}{\sin }^{2n+1}x\mathrm{d}x\le {\left({\int }_0^{\frac{\pi }{2}}{\sin }^{2n}x\mathrm{d}x\right)}^2\le {\int }_0^{\frac{\pi }{2}}{\sin }^{2n}x\mathrm{d}x{\int }_0^{\frac{\pi }{2}}{\sin }^{2n-1}x\mathrm{d}x\\ \Rightarrow & \frac{(2n-1)!!}{(2n)!!}\frac{(2n)!!}{(2n+1)!!}\frac{\pi }{2}\le {\left({\int }_0^{\frac{\pi }{2}}{\sin }^{2n}x\mathrm{d}x\right)}^2\le \frac{(2n-1)!!}{(2n)!!}\frac{(2n-2)!!}{(2n-1)!!}\frac{\pi }{2}\\ & \\ \Rightarrow & \frac{\pi }{2}\frac{1}{2n+1}\le {\left(\frac{(2n-1)!!}{(2n)!!}\right)}^2\le \frac{1}{2n}\frac{\pi }{2}\end{array}$$

So

$$\underset{n\to \infty }{\lim }\sqrt{n}\frac{(2n-1)!!}{(2n)!!}=\frac{\sqrt{\pi }}{2}$$

ExampleCaulate

$$\underset{n\to \infty }{\lim }\sum _{i=0}^n{\lambda }^i{a}_{n-i}$$

Solve

$$\begin{array}{rl}\underset{n\to \infty }{\lim }\sum _{i=0}^n{\lambda }^i{a}_{n-i}& =\underset{n\to \infty }{\lim }\sum _{i=0}^n{\lambda }^{n-i}{a}_i\\ & =\underset{n\to \infty }{\lim }{\lambda }^n\sum _{i=0}^n\frac{a_i}{{\lambda }^i}\\ & =\underset{n\to \infty }{\lim }\frac{{\sum }_{i=0}^n\frac{a_i}{{\lambda }^i}}{{\left(\frac{1}{\lambda }\right)}^n}\\ & =\underset{n\to \infty }{\lim }\frac{\frac{a_n}{{\lambda }^n}}{{\left(\frac{1}{\lambda }\right)}^{n-1}\left(\frac{1}{\lambda }-1\right)}\\ & =\frac{a}{\lambda -1}\end{array}$$

Example Please use Cauchy Convergence Principle prove :A monotonic and bounded sequence must converge.

Proof

Only prove the case where an increasing sequence bounded above must have a limit.

Let’s assume without loss of generality that {xn}\{x_n\}{xn​} is an increasing sequence bounded above.

We will use the method of proof by contradiction to show that {xn}\{x_n\}{xn​} has a limit.

Suppose that the sequence {xn}\{x_n\}{xn​} does not have a limit. According to the Cauchy convergence criterion, there exists ${\epsilon }_0>0$ such that for all $N$, there exist $n>N$ where

$$|x_n-x_N|=x_n-x_N>{\epsilon }_0$$

Successively take $N_1=1$, then there exists $n_1>N_1$ such that $x_{n_1}-x_1>{\epsilon }_0$; $N_2=n_1$, then there exists $n_2>n_1$ such that

$$\begin{array}{r}{x}_{n_2}-x_{n_1}>{\epsilon }_0\\ ⋮\\ N_k=n_{k-1}\end{array}$$
then there exists $n_k>n_{k-1}$ such that $x_{n_k}-x_{n_{k-1}}>{\epsilon }_0$. Adding up the above - listed inequalities, we get $x_{n_k}-x_1>k{\epsilon }_0$. For any real number $G$, when $k>\frac{G-x_1}{{\epsilon }_0}$, we have $x_{n_k}>G$. This contradicts the fact that {xn}\{x_n\}{xn​} is bounded above.

Therefore, the sequence {xn}\{x_n\}{xn​} has a limit.

Example
Let $A_n={\sum }_{k=1}^n{a}_k$, and $\underset{n\to \infty }{\lim }{A}_n$ exists. {pn}\{p_n\}{pn​} is a sequence of positive numbers that is monotonically increasing, and $p_n\to +\infty$ as $n\to \infty$. Prove that:

$$\underset{n\to \infty }{\lim }\frac{p_1{a}_1+p_2{a}_2+\cdots +p_n{a}_n}{p_n}=0$$

Prove

$$\begin{array}{rl}\frac{{\sum }_{i=1}^n{p}_i{a}_i}{p_n}& =\frac{{\sum }_{i=2}^n{p}_i(A_i-A_{i-1})+p_1{A}_1}{p_n}\\ & =\frac{{\sum }_{i=1}^{n-1}(p_i-p_{i+1})A_i+A_n{p}_n}{p_n}\end{array}$$

So

$$\begin{array}{rl}\underset{n\to \infty }{\lim }\frac{{\sum }_{i=1}^n{p}_i{a}_i}{p_n}& =\underset{n\to \infty }{\lim }\frac{{\sum }_{i=1}^{n-1}(p_i-p_{i+1})A_i+A_n{p}_n}{p_n}\\ & =-A+A=0\end{array}$$

Continuity and Limits of Functions

Example Let$f(x)$satisfy the functional equation $f(2x)=f(x)$ on $(0,+\infty )$, and
$\underset{x\to +\infty }{\lim }f(x)=A$ .Prove that
$$f(x)\equiv A$$
for $x\in (0,+\infty )$.

Prove

$\forall x_0\in (0,+\infty )$,we have
$$f(x_0)=f(2x_0)=\cdots =f(2^n{x}_0)$$
Let $n\to \infty$
$$f(x_0)=\underset{n\to \infty }{\lim }f(2^n{x}_0)=f(\underset{n\to \infty }{\lim }{2}^n{x}_0)=f(+\infty )=A$$

By arbitrariness of $x_0$,we have:

$$f(x)\equiv A$$

DefinitionnLet $f:I\to \mathbb{R}$ be a function. $f$ is said to be uniformly continuous on $I$ if for every $\epsilon >0$, there exists a $\delta >0$ such that for all $x_1,x_2\in I$ with $|x_1-x_2|<\delta$, we have $|f(x_1)-f(x_2)|<\epsilon$.

ExampleIf the function $f(x)$ is continuous on $[a,+\infty )$ and
$$\underset{x\to +\infty }{\lim }f(x)=A$$
then $f(x)$is uniformly continuous on $[a,+\infty )$.

Prove

Since $\underset{x\to +\infty }{\lim }f(x)=A$, for any $\epsilon >0$, there exists $X>a$ such that for all $x^{\prime },x^{\prime \prime }>X$, the inequality $|f(x^{\prime })-f(x^{\prime \prime })|<\epsilon$ holds. Because $f(x)$ is continuous on $[a,X+1]$, it is uniformly continuous. That is, there exists $0<\delta <1$ such that for all $x^{\prime },x^{\prime \prime }\in [a,X+1]$, when $|x^{\prime }-x^{\prime \prime }|<\delta$, $|f(x^{\prime })-f(x^{\prime \prime })|<\epsilon$. Then, for all $x^{\prime },x^{\prime \prime }\in [a,+\infty )$, when $|x^{\prime }-x^{\prime \prime }|<\delta$, $|f(x^{\prime })-f(x^{\prime \prime })|<\epsilon$.

ExampleLet $f(x)$ be continuous on $[a,b]$, $f(a)=f(b)=0$, and $f_{+}^{\prime }(a)\cdot f_{-}^{\prime }(b)>0$. Prove that $f(x)$ has at least one zero point in $(a,b)$.

Prove

• $f_{+}^{\prime }(a)$: The right - hand derivative of $f(x)$ at $x=a$, defined as $f_{+}^{\prime }(a)=\underset{x\to a^{+}}{\lim }\frac{f(x)-f(a)}{x-a}=\underset{x\to a^{+}}{\lim }\frac{f(x)}{x-a}$ (since $f(a)=0$).
• $f_{-}^{\prime }(b)$: The left - hand derivative of $f(x)$ at $x=b$, defined as $f_{-}^{\prime }(b)=\underset{x\to b^{-}}{\lim }\frac{f(x)-f(b)}{x-b}=\underset{x\to b^{-}}{\lim }\frac{f(x)}{x-b}$ (since $f(b)=0$).

Since $f_{+}^{\prime }(a)\cdot f_{-}^{\prime }(b)>0$, we can know that $f_{+}^{\prime }(a)$ and $f_{-}^{\prime }(b)$ have the same sign. Then, by the definition of one - sided derivatives, we can find two points $x_1\in (a,a+{\delta }_1)$ and $x_2\in (b-{\delta }_2,b)$ such that $f(x_1)$ and $f(x_2)$ have opposite signs . Then, by the Intermediate Value Theorem, there must be a zero point of $f(x)$ in the interval $(x_1,x_2)\subseteq (a,b)$.

Example
$${\left(x^{n-1}{e}^{\frac{1}{x}}\right)}^{(n)}=\frac{(-1{)}^n}{x^{n+1}}{e}^{\frac{1}{x}}$$

By Faà di Bruno’s Formula:

$$\frac{{\mathrm{d}}^n}{\mathrm{d}{x}^n}f(g(x))=\sum \frac{n!}{m_1!m_2!\cdots m_n!}\cdot f^{(m_1+m_2+\cdots +m_n)}(g(x))\cdot \prod _{j=1}^n{\left(\frac{g^{(j)}(x)}{j!}\right)}^{m_j}$$

where:
The summation $\sum$ ranges over all sets of non - negative integers $m_1,m_2,\dots ,m_n$ satisfying:
$$1\cdot m_1+2\cdot m_2+3\cdot m_3+\cdots +n\cdot m_n=n$$

$$\begin{array}{rl}{\left(x^{n-1}{e}^{\frac{1}{x}}\right)}^{(n)}& =e^{\frac{1}{x}}\sum _{k=0}^{n-1}\left(\genfrac{}{}{0ex}{}{n}{k}\right)\frac{(n-1)!}{(n-1-k)!}{x}^{n-1-k}\sum _{\begin{array}{c}1\cdot m_1+\cdots +(n-k)\cdot m_{n-k}=n-k\end{array}}\frac{(n-k)!}{m_1!\cdots m_{n-k}!}\prod _{j=1}^{n-k}{\left[\frac{(-1{)}^j}{x^{j+1}}\right]}^{m_j}\\ & \stackrel{M=m_1+\cdots +m_{n-k}}{=}{e}^{\frac{1}{x}}\sum _{k=0}^{n-1}\left(\genfrac{}{}{0ex}{}{n}{k}\right)\frac{(n-1)!}{(n-1-k)!}{x}^{n-1-k}\cdot (-1{)}^{n-k}\cdot \sum \frac{(n-k)!}{m_1!\cdots m_{n-k}!}\cdot x^{-((n-k)+M)}\\ & \stackrel{\text{唯一有效项为}k=0}{=}{x}^{n-1}\cdot (-1{)}^n{e}^{\frac{1}{x}}\cdot x^{-2n}=\frac{(-1{)}^n}{x^{n+1}}{e}^{\frac{1}{x}}\end{array}$$

Mathematical induction is easy

# Advanced  Math & Math Analysis |01 Limits, Continuous `Note: For simple definitions and corollaries, we will not elaborate.`@[toc]
## Some easy knowledge### Mathematical induction>**Theorem**
Let $T(n)$ is a proposition if:
>
>(1) $T(1)$ is true;
>
>(2)let $T(k)(k\ge 1)$ is true;
>
>(3) prove: $T(k+1)$is true.
>
>So $T(n)$ is true.**Example**Let $n\ge 1$ and $n\in \mathbb{N}$,prove $$n!< \left(\frac{n+1}{2}\right)^n$$**Proof**When $n = 2$, $2! = 2 < \frac{9}{4} = \left( \frac{2 + 1}{2} \right)^2$, and the proposition holds.Suppose the proposition holds when $n = k (k \geq 2)$, that is, $$ k! < \left( \frac{k + 1}{2} \right)^k $$When $n = k + 1$, $$ \begin{align*} (k + 1)!&= k!(k + 1)\\ &< \left( \frac{k + 2}{2} \cdot \frac{k + 1}{k + 2} \right)^k \cdot (k + 1)\\ &= \left( \frac{k + 2}{2} \right)^k \cdot \frac{(k + 1)^{k + 1}}{(k + 2)^k}\\ &= \frac{(k + 2)^{k + 1}}{2^k} \cdot \left( \frac{k + 1}{k + 2} \right)^{k + 1}\\ &= 2 \left( \frac{k + 2}{2} \right)^{k + 1} \cdot \frac{1}{\left( \frac{k + 2}{k + 1} \right)^{k + 1}}\\ &= 2 \left( \frac{k + 2}{2} \right)^{k + 1} \cdot \frac{1}{\left( 1 + \frac{1}{k + 1} \right)^{k + 1}} \end{align*} $$ By $(1+x)^n\ge 1+nx$,$x\in(-1,+\infty)$,$n\in \mathbb{N}_+$, we have $$ \left( 1 + \frac{1}{k + 1} \right)^{k + 1} > 1 + (k + 1) \cdot \frac{1}{k + 1} = 2 $$ So $$ (k + 1)! < 2 \left( \frac{k + 2}{2} \right)^{k + 1} \cdot \frac{1}{2} = \left( \frac{k + 2}{2} \right)^{k + 1} $$Thus, when $n = k + 1$, the proposition holds. The proof is completed. ### Some equation of trigonometric function### Inequality## Limits of sequence>**Example** Prove  an open interval has no maximum value.**Proof**For any open continue interval $(a,b)$ can be mapping to $(0,1)$ , by$$f:\frac{x-a}{b-a}$$Suppose there exists a  $\max \{(0,1)\} \stackrel{\scriptstyle\triangle}{=} \alpha $ ,we have$$
\begin{align*}&0< \frac{1}{2}\alpha <\frac{1}{2} \\&\Rightarrow  \frac{1}{2} < \frac{\alpha +1}{2} <1
\end{align*}
$$Obviously,$$\alpha < \frac{\alpha +1}{2}$$This contradicts the assumption, so the assumption is wrong.>**Example** Let $T = \{ x \mid x \in \mathbb{Q} \text{ and } x > 0, x^2 < 2 \}$. Prove that $T$ has no upper bound within $\mathbb{Q}$.**Proof** Use proof by contradiction. Suppose $T$ has a least upper bound within $\mathbb{Q}$. Denote $\sup T = \frac{n}{m}$ (where $m, n \in \mathbb{N}^*$ and $m, n$ are coprime). Then, it is obvious that: $$ 1 < \left( \frac{n}{m} \right)^2 < 3 $$Since the square of a rational number cannot equal 2, there are only the following two possibilities: (1) $1 < \left( \frac{n}{m} \right)^2 < 2$: Let $2 - \frac{n^2}{m^2} = t$, then $0 < t < 1$. Let $r = \frac{n}{6m^2}t$. Obviously, $\frac{n}{m} + r > 0$, and $\frac{n}{m} + r \in \mathbb{Q}$. Since $r^2 = \frac{n^2}{36m^4}t^2 < \frac{1}{18}t$, and $\frac{2n}{m}r = \frac{n^2}{3m^3}t < \frac{2}{3}t$, we can obtain: 
$$ \left( \frac{n}{m} + r \right)^2 - 2 = r^2 + \frac{2n}{m}r - t < 0 $$
This shows that $\frac{n}{m} + r \in T$, which contradicts the fact that $\frac{n}{m}$ is the least upper bound of $T$. (2) $2 < \left( \frac{n}{m} \right)^2 < 3$: Let $\frac{n^2}{m^2} - 2 = t$, then $0 < t < 1$. Let $r = \frac{n}{6m^2}t$. Obviously, $\frac{n}{m} - r > 0$, and $\frac{n}{m} - r \in \mathbb{Q}$. Since $\frac{2n}{m}r = \frac{n^2}{3m^3}t < t$, we can obtain: 
$$ \left( \frac{n}{m} - r \right)^2 - 2 = r^2 - \frac{2n}{m}r + t > 0 $$
This shows that $\frac{n}{m} - r$ is also an upper bound of $T$, which contradicts the fact that $\frac{n}{m}$ is the least upper bound of $T$. >**Example** Caculate
>
>$$\lim_{n\to \infty} \frac{(2n-1)!!}{(2n)!!}$$
>and
>$$\lim_{n\to \infty}\sqrt{n} \frac{(2n-1)!!}{(2n)!!}$$**Solve**(1)$$\boxed{\lim_{n\to \infty} \frac{(2n-1)!!}{(2n)!!}=0}$$$$\begin{align*}
0\leq \lim_{n\to \infty} \frac{(2n-1)!!}{(2n)!!}&=\lim_{n\to \infty} \frac{\prod_{i=1}^n (2i-1)}{\prod_{i=1}^n (2i)}\\
&=\sqrt{\lim_{n\to \infty}\left( \frac{\prod_{i=1}^n (2i-1)}{\prod_{i=1}^n (2i)}\right)^2}
\\&\leq \sqrt{\lim_{n\to \infty} \frac{\left(\prod_{i=1}^n (2i-1)\right)^2}{\prod_{i=1}^n (2i+1)\prod_{i=1}^n (2i-1)}}\\
&=\sqrt{\lim_{n\to \infty} \frac{\prod_{i=1}^n (2i-1)}{\prod_{i=1}^n (2i+1)}}\\&=\sqrt{\lim_{n\to \infty} \frac{1}{2n+1}}
\\&=0
\end{align*}$$(2)$$\boxed{\lim_{n\to \infty}\sqrt{n} \frac{(2n-1)!!}{(2n)!!}=\frac{\sqrt{\pi}}{2}}$$$$
\begin{align*}
&\int_{0}^{\frac{\pi}{2}}\sin^{2n} x\mathrm{d}x \int_{0}^{\frac{\pi}{2}}\sin^{2n+1} x\mathrm{d}x \leq \left(\int_{0}^{\frac{\pi}{2}}\sin^{2n} x\mathrm{d}x\right)
^2 \leq \int_{0}^{\frac{\pi}{2}}\sin^{2n} x\mathrm{d}x\int_{0}^{\frac{\pi}{2}}\sin^{2n-1} x\mathrm{d}x
\\\Rightarrow  & \frac{(2n-1)!!}{(2n)!!}\frac{(2n)!!}{(2n+1)!!}\frac{\pi}{2} \leq \left(\int_{0}^{\frac{\pi}{2}}\sin^{2n} x\mathrm{d}x\right)^2 \leq \frac{(2n-1)!!}{(2n)!!}\frac{(2n-2)!!}{(2n-1)!!}\frac{\pi}{2} \\
\\\Rightarrow  &  \frac{\pi}{2} \frac{1}{2n+1}\leq \left(\frac{(2n-1)!!}{(2n)!!}\right)^2 \leq \frac{1}{2n}\frac{\pi}{2} 
\end{align*}
$$So$$\lim_{n\to \infty}\sqrt{n} \frac{(2n-1)!!}{(2n)!!}=\frac{\sqrt{\pi}}{2}$$> **Example**Caulate
>
>$$\lim_{n \to \infty}\sum_{i=0}^n \lambda ^i a_{n-i}$$**Solve**$$
\begin{align*}
\lim_{n \to \infty}\sum_{i=0}^n \lambda ^i a_{n-i}&=\lim_{n \to \infty}\sum_{i=0}^n \lambda ^{n-i} a_{i}\\
&=\lim_{n \to \infty}\lambda ^{n}\sum_{i=0}^n \frac{ a_{i}}{\lambda ^{i}}\\&=\lim_{n \to \infty}\frac{\sum_{i=0}^n \frac{ a_{i}}{\lambda ^{i}}}{\left(\frac{1}{\lambda}\right) ^{n}}
\\
&=\lim_{n \to \infty}\frac{ \frac{ a_{n}}{\lambda ^{n}}}{\left(\frac{1}{\lambda}\right) ^{n-1}\left(\frac{1}{\lambda}-1\right)}\\
&=\frac{a}{\lambda -1}
\end{align*}
$$>**Example** Please use Cauchy Convergence Principle prove :A monotonic and bounded sequence must converge.**Proof**Only prove the case where an increasing sequence bounded above must have a limit. Let's assume without loss of generality that $\{x_n\}$ is an increasing sequence bounded above. We will use the method of proof by contradiction to show that $\{x_n\}$ has a limit. Suppose that the sequence $\{x_n\}$ does not have a limit. According to the Cauchy convergence criterion, there exists $\varepsilon_0> 0$ such that for all $N$, there exist $n > N$ where $$|x_n - x_N|=x_n - x_N>\varepsilon_0$$Successively take $N_1 = 1$, then there exists $n_1>N_1$ such that $x_{n_1}-x_1 >\varepsilon_0$; $N_2 = n_1$, then there exists $n_2 > n_1$ such that $$
\begin{align*}
x_{n_2}-x_{n_1}>\varepsilon_0\\ \vdots \\
N_k=n_{k - 1}\\
\end{align*}$$
then there exists $n_k>n_{k - 1}$ such that $x_{n_k}-x_{n_{k - 1}}>\varepsilon_0$. Adding up the above - listed inequalities, we get $x_{n_k}-x_1>k\varepsilon_0$. For any real number $G$, when $k>\frac{G - x_1}{\varepsilon_0}$, we have $x_{n_k}>G$. This contradicts the fact that $\{x_n\}$ is bounded above. Therefore, the sequence $\{x_n\}$ has a limit. >**Example**Let $A_n=\sum_{k = 1}^{n}a_k$, and $\lim\limits_{n \to \infty}A_n$ exists. $\{p_n\}$ is a sequence of positive numbers that is monotonically increasing, and $p_n\to+\infty$ as $n\to\infty$. Prove that:
>
>$$
\lim_{n \to \infty}\frac{p_1a_1 + p_2a_2+\cdots + p_na_n}{p_n}=0
$$  **Prove**$$
\begin{align*}
\frac{\sum_{i=1}^np_ia_i}{p_n}&=\frac{\sum_{i=2}^np_i(A_i-A_{i-1})+p_1A_1}{p_n}\\
&=\frac{\sum_{i=1}^{n-1}(p_i-p_{i+1})A_i +A_np_n}{p_n}
\end{align*}
$$So$$
\begin{align*}
\lim\limits_{n \to \infty}\frac{\sum_{i=1}^np_ia_i}{p_n}
&=\lim\limits_{n \to \infty}\frac{\sum_{i=1}^{n-1}(p_i-p_{i+1})A_i +A_np_n}{p_n}\\
&=-A+A=0
\end{align*}$$>**Example**15. Let$f(x)$satisfy the functional equation $f(2x)=f(x)$ on $(0, +\infty)$, and 
$\lim\limits_{x \to +\infty}f(x) = A$ .Prove that 
>$$f(x)\equiv A$$
>for $x\in(0, +\infty)$. **Prove**$\forall x_0 \in (0,+\infty)$,we have
$$f(x_0)=f(2x_0)=\cdots=f(2^nx_0)$$
Let $n \to \infty$
$$f(x_0)=\lim_{n\to \infty}f(2^nx_0)=f(\lim_{n\to \infty}2^nx_0)=f(+\infty)=A$$By arbitrariness of $x_0$,we have:$$f(x)\equiv A $$>**Definitionn**Let $f:I\rightarrow\mathbb{R}$ be a function. $f$ is said to be **uniformly continuous** on $I$ if for every $\varepsilon > 0$, there exists a $\delta>0$ such that for all $x_1,x_2\in I$ with $|x_1 - x_2|<\delta$, we have $|f(x_1)-f(x_2)|<\varepsilon$.>**Example**If the function $f(x)$ is continuous on $[a, +\infty)$ and 
>$$\lim\limits_{x \to +\infty} f(x) = A
>$$ 
>then $f(x)$is uniformly continuous on $[a, +\infty)$.  **Prove**Since $\lim\limits_{x \to +\infty} f(x)=A$, for any $\varepsilon> 0$, there exists $X > a$ such that for all $x', x'' > X$, the inequality $|f(x') - f(x'')| < \varepsilon$ holds.  Because $f(x)$ is continuous on $[a, X + 1]$, it is uniformly continuous. That is, there exists $0 < \delta < 1$ such that for all $x', x'' \in [a, X + 1]$, when $|x' - x''| < \delta$,  $|f(x') - f(x'')| < \varepsilon$.  Then, for all $x', x'' \in [a, +\infty)$, when $|x' - x''| < \delta$,  $|f(x') - f(x'')| < \varepsilon$. 
>**Example**Let $f(x)$ be continuous on $[a, b]$, $f(a)=f(b) = 0$, and $f'_{+}(a)\cdot f'_{-}(b)>0$. Prove that $f(x)$ has at least one zero point in $(a, b)$.**Prove**- $f'_{+}(a)$: The right - hand derivative of $f(x)$ at $x = a$, defined as $f'_{+}(a)=\lim\limits_{x\rightarrow a^{+}}\frac{f(x)-f(a)}{x - a}=\lim\limits_{x\rightarrow a^{+}}\frac{f(x)}{x - a}$ (since $f(a) = 0$).
- $f'_{-}(b)$: The left - hand derivative of $f(x)$ at $x = b$, defined as $f'_{-}(b)=\lim\limits_{x\rightarrow b^{-}}\frac{f(x)-f(b)}{x - b}=\lim\limits_{x\rightarrow b^{-}}\frac{f(x)}{x - b}$ (since $f(b)=0$). Since $f'_{+}(a)\cdot f'_{-}(b)>0$, we can know that $f'_{+}(a)$ and $f'_{-}(b)$ have the same sign. Then, by the definition of one - sided derivatives, we can find two points $x_1\in(a,a + \delta_1)$ and $x_2\in(b-\delta_2,b)$ such that $f(x_1)$ and $f(x_2)$ have opposite signs . Then, by the Intermediate Value Theorem, there must be a zero point of $f(x)$ in the interval $(x_1,x_2)\subseteq(a,b)$. >**Example**
>$$\left(x^{n-1}e^{\frac{1}{x}}\right)^{(n)}=\frac{(-1)^n}{x^{n+1}}e^{\frac{1}{x}}$$By **Faà di Bruno's Formula**:  $$
\frac{\mathrm{d}^n}{\mathrm{d}x^n} f(g(x)) = \sum \frac{n!}{m_1! \, m_2! \, \cdots \, m_n!} \cdot f^{(m_1 + m_2 + \cdots + m_n)}(g(x)) \cdot \prod_{j=1}^n \left( \frac{g^{(j)}(x)}{j!} \right)^{m_j}
$$  where:  
The summation $\sum$ ranges over all sets of non - negative integers $m_1, m_2, \dots, m_n$ satisfying:  
$$1 \cdot m_1 + 2 \cdot m_2 + 3 \cdot m_3 + \cdots + n \cdot m_n = n$$  $$
\begin{align*}
\left(x^{n-1}e^{\frac{1}{x}}\right)^{(n)}&=e^{\frac{1}{x}}\sum_{k=0}^{n-1}\binom{n}{k}\frac{(n-1)!}{(n-1-k)!} x^{n-1-k}\sum_{\substack{1 \cdot m_1 + \cdots + (n-k) \cdot m_{n-k} = n-k}}\frac{(n-k)!}{m_1!\cdots m_{n-k}!}\prod_{j=1}^{n-k}\left[\frac{(-1)^j}{x^{j+1}}\right]^{m_j}\\
&\stackrel{M=m_1+\cdots +m_{n-k}}{=}e^{\frac{1}{x}}\sum_{k=0}^{n-1}\binom{n}{k}\frac{(n-1)!}{(n-1-k)!} x^{n-1-k}\cdot (-1)^{n-k} \cdot \sum \frac{(n-k)!}{m_1! \cdots m_{n-k}!} \cdot x^{-((n-k) + M)}\\
&\stackrel{\text{唯一有效项为}k = 0}{=}x^{n-1} \cdot (-1)^n e^{\frac{1}{x}} \cdot x^{-2n} = \frac{(-1)^n}{x^{n+1}} e^{\frac{1}{x}}
\end{align*}
$$`Mathematical induction is easy`