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# 数学代写|实分析代写Real Analysis代考|Existence of Radius of Convergence

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## 数学代写|实分析代写Real Analysis代考|Existence of Radius of Convergence

We consider the sequence $S=\left(\left|a_1\right|, \sqrt{\left|a_2\right|}, \sqrt[3]{\left|a_3\right|}, \sqrt[4]{\left|a_4\right|}, \ldots\right)$. If this sequence is unbounded, then for every $x \neq 0$, the sequence $\left(\left|a_1\right||x|, \sqrt{\left|a_2\right|}|x|, \sqrt[3]{\left|a_3\right|}|x|, \sqrt[4]{\left|a_4\right|}|x|, \ldots\right)$ is also unbounded. By the root test (Theorem 4.9), the power series diverges at every value of $x$ other than $x=0$. In this case, the radius of convergence is zero. If $S$ is bounded, then $\varlimsup_{n \rightarrow \infty} \sqrt[n]{\left|a_n\right|}$ will always be well defined and greater than or equal to zero. It still remains for us to prove that $R=1 / \varlimsup_{n \rightarrow \infty} \sqrt[n]{\left|a_n\right|}$ is a radius of convergence.

Theorem 4.14 (Existence of Radius of Convergence). Let $a_0+\sum_{n=1}^{\infty} a_n x^n$ be an arbitrary power series and define
$$R=\frac{1}{\overline{\lim }{n \rightarrow \infty} \sqrt[n]{\left|a_n\right|}}$$ This series converges absolutely for $|x|R$. The power series converges at all values of $x$ when $\overline{\lim }{n \rightarrow \infty} \sqrt[n]{\left|a_n\right|}=0$, and it converges only at $x=0$ when the upper limit is infinite.

Proof: Let $\lambda=\varlimsup_{n \rightarrow \infty} \sqrt[n]{\left|a_n\right|}$. If $|x|<1 / \lambda$, then we can find an $\alpha$ just a little less than 1 and an $\epsilon$ just a little larger than zero so that we still have $$|x|<\frac{\alpha}{\lambda+\epsilon} .$$ It follows that $$\sqrt[n]{\left|a_n x^n\right|}=\sqrt[n]{\left|a_n\right|}|x|<\frac{\sqrt[n]{\left|a_n\right|}}{\lambda+\epsilon} \alpha .$$ By the definition of $\lambda$ as the upper limit of $\sqrt[n]{\left|a_n\right|}$, this last term is strictly less than $\alpha$ for all sufficiently large values of $n$. The root test, Theorem 4.9 , tells us that the series converges absolutely. If $|x|>1 / \lambda$, then we can find an $\epsilon$ just a little larger than zero so that we still have
$$|x|>\frac{1}{\lambda-\epsilon} .$$
It follows that
$$\sqrt[n]{\left|a_n x^n\right|}=\sqrt[n]{\left|a_n\right|}|x|>\frac{\sqrt[n]{\left|a_n\right|}}{\lambda-\epsilon}$$

## 数学代写|实分析代写Real Analysis代考|Hypergeometric Series

What happens when $|x|$ equals the radius of convergence? The series might converge at both endpoints, diverge at both, or converge at only one of these values. If it converges at both, the convergence might be absolute or conditional. There is no single test that will return a conclusive answer for all power series, but in 1812 Carl Friedrich Gauss did publish a test that determines the convergence at the endpoints for every power series you are likely to encounter outside of a course in real analysis. It is a definitive test that works when the power series is hypergeometric.
The easiest infinite series with which to work is the geometric series,
$$1+x+x^2+x^3+\cdots .$$

It converges to $1 /(1-x)$ when $|x|<1$, and it diverges when $|x| \geq 1$. In the seventeenth and eighteenth centuries, mathematicians began to appreciate a larger class of series that was almost as nice, the hypergeometric series. A geometric series is characterized by the fact that the ratio of two successive summands is constant. In a hypergeometric series, the ratio of two succesive nonzero summands is a rational function of the subscript.
Definition: hypergeometric series
A series $a_1+a_2+a_3+\cdots$ is hypergeometric if
$$\frac{a_{n+1}}{a_n}=\frac{P(n)}{Q(n)}$$
where $P(n)$ and $Q(n)$ are polynomials in $n$.
For example, the exponential series is hypergeometric:
$$\frac{a_{n+1}}{a_n}=\frac{x^n / n !}{x^{n-1} /(n-1) !}=\frac{x}{n} .$$
The numerator is the constant $x$ (constant with respect to $n$ ), and the denominator is the linear function $n$. The series for $\sin x$ is also hypergeometric:
$$\frac{a_{n+1}}{a_n}=\frac{(-1)^n x^{2 n+1} /(2 n+1) !}{(-1)^{n-1} x^{2 n-1} /(2 n-1) !}=\frac{-x^2}{(2 n)(2 n+1)}$$

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## 数学代写|实分析代写Real Analysis代考|Existence of Radius of Convergence

$$R=\frac{1}{\overline{\lim } n \rightarrow \infty \sqrt[n]{\left|a_n\right|}}$$

$$|x|>\frac{1}{\lambda-\epsilon}$$

$$\sqrt[n]{\left|a_n x^n\right|}=\sqrt[n]{\left|a_n\right|} x \mid>\frac{\sqrt[n]{\left|a_n\right|}}{\lambda-\epsilon}$$

## 数学代写|实分析代写Real Analysis代考|Hypergeometric Series

$$1+x+x^2+x^3+\cdots$$

A级数 $a_1+a_2+a_3+\cdots$ 是超几何的，如果
$$\frac{a_{n+1}}{a_n}=\frac{P(n)}{Q(n)}$$

$$\frac{a_{n+1}}{a_n}=\frac{x^n / n !}{x^{n-1} /(n-1) !}=\frac{x}{n} .$$

$$\frac{a_{n+1}}{a_n}=\frac{(-1)^n x^{2 n+1} /(2 n+1) !}{(-1)^{n-1} x^{2 n-1} /(2 n-1) !}=\frac{-x^2}{(2 n)(2 n+1)}$$

## MATLAB代写

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