1.10

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Power of Series:

$\sum_{n=0}^\infty a_n x^n$ is centered at $x=0$.

$\{a_n\}$ is a sequence.

$\sum_{n=0}^\infty a_n (x-a)^n$ is centered at $x=a$.

Convergent set: $CS\{a_n\} = \{ x \in \mathbb{R} : \sum_{n=0}^\infty a_n x^n \text{ converges} \}$.

Absolutely convergent set: $AS\{a_n\} = \{ x \in \mathbb{R} : \sum_{n=0}^\infty |a_n x^n| \text{ converges} \}$.

$AS\{a_n\} \subseteq CS\{a_n\}$.

Convergence Radius $R$: $f(x) = \sum_{n=0}^\infty a_n x^n$ is well-defined in $(-R, R)$, and $f$ is continuous.

Examples:

$\sum_{n=1}^\infty \frac{x^n}{n!} = e^x$

$\sum_{n=1}^\infty \left| \frac{x^n}{n!} \right|$

D'Alembert criterion: $\lim_{n\to\infty}\left|\frac{\frac{x^{n+1}}{(n+1)!}}{\frac{x^{n}}{n!}}\right|=\lim _{n\to\infty}\left|x\cdot\frac{1}{n+1}\right|=|x|\cdot\lim_{n\to\infty}\frac{1}{n+1} <1$

Then $\sum_{n=1}^\infty \frac{x^n}{n!}$ converges for all $x$. Thus $R = \infty$
$\sum_{n=1}^\infty \frac{x^n}{3^n \sqrt{n}}$

$\sum_{n=1}^\infty \left| \frac{x^n}{3^n \sqrt{n}} \right|$

Using D'Alembert's criterion:

$\lim_{n\to\infty}\left|\frac{\frac{x^{n+1}}{3^{n+1}\sqrt{n+1}}}{\frac{x^{n}}{3^{n}\sqrt{n}}} \right|=\lim_{n\to\infty}\left|\frac{|x|}{3}\cdot\sqrt{\frac{n}{n+1}}\right|=\frac{|x|}{3}\cdot\lim_{n\to\infty}\sqrt{\frac{n}{n+1}}<\frac{|x|}{3}\cdot1=\frac{|x|}{3} $

For convergence: $\frac{|x|}{3} < 1 \implies |x| < 3.$

Thus, $R = 3$, and the radius of convergence is $3$.

For $x = -3$: $\sum_{n=1}^\infty \frac{(-3)^n}{3^n \sqrt{n}}= \sum_{n=1}^\infty \frac{(-1)^n}{\sqrt{n}}$

This series is conditionally convergent.

For $x = 3$: $\sum_{n=1}^\infty \frac{(3)^n}{3^n \sqrt{n}}= \sum_{n=1}^\infty \frac{1}{\sqrt{n}}$

This is a divergent series.

Convergent set (CS): $[-3, 3)$.

Absolutely convergent set (ACS): $(-3, 3)$.
$\sum_{n=0}^\infty n! x^n = \sum_{n=0}^\infty n! |x|^n$

Using D'Alembert's criterion: $\lim_{n\to\infty}\frac{(n+1)!|x|^{n+1}}{n!|x|^{n}}=\lim_{n\to\infty}(n+1)|x|=|x| \cdot\underbrace{\lim_{n\to\infty}(n+1)}_{r}<1$

Then $|x|<\frac{1}{r}\rightarrow 0$, then $|x|\leq 0$, then $R=0$
$\sum_{n=1}^\infty \frac{(-1)^{n+1} x^n}{n}$

Using D'Alembert's criterion:

$\lim_{n\to\infty}\left|\frac{\frac{(-1)^{n+2}x^{n+1}}{n+1}}{\frac{(-1)^{n+1}x^{n}}{n}} \right|=\lim_{n\to\infty}\left|x\cdot\frac{n}{n+1}\right|=|x|\cdot\lim_{n\to\infty} \frac{n}{n+1}<1$

For convergence: $|x| < 1$.

Convergent set (CS): $(-1, 1]$

Absolutely convergent set (ACS): $(-1, 1)$

Because

At $x = -1$: $\sum_{n=1}^{\infty}\frac{(-1)^{n+1}(-1)^{n}}{n}=\sum_{n=1}^{\infty}\frac{-1}{n}= -\sum_{n=1}^{\infty}\frac{1}{n}$ which is divergent.

At $x = 1$: $\sum_{n=1}^{\infty}\frac{(-1)^{n+1}(1)^{n}}{n}=\sum_{n=1}^{\infty}\frac{(-1)^{n+1}}{n}$. This series is conditionally convergent.
$\sum_{n=1}^\infty \frac{3^n x^n}{n^n}$

Using D'Alembert's criterion: $\lim_{n\to\infty}\left|\frac{\frac{3^{n+1}x^{n+1}}{(n+1)^{n+1}}}{\frac{3^{n}x^{n}}{n^{n}}} \right|=3|x|\cdot\lim_{n\to\infty}\left(\frac{n}{n+1}\right)^{n}\cdot\frac{1}{n+1}$

$= 3|x| \cdot \lim_{n \to \infty} \left( \frac{1}{\frac{n+1}{n}} \right)^n \cdot \lim_{n \to \infty} \frac{1}{n+1} = 3|x| \cdot \lim_{n \to \infty} \left( \frac{1}{1 + \frac{1}{n}} \right)^n \cdot \lim_{n \to \infty} \frac{1}{n+1} = 3|x| \cdot \frac{1}{e} \cdot 0 = \frac{3|x|}{e} \cdot 0 < 1.$

Thus, $R = \infty$.
$\sum_{n=1}^\infty \frac{1}{n + \sqrt{n}} \cdot x^n$

Using D'Alembert's criterion: $\lim_{n\to\infty}\left|\frac{\frac{x^{n+1}}{n+1+\sqrt{n+1}}}{\frac{x^{n}}{n+\sqrt{n}}} \right|=\lim_{n\to\infty}\left|x\cdot\frac{n+\sqrt{n}}{n+1+\sqrt{n+1}}\right|$

Use comparasion test, we get it is $=|x|<1$

$R = 1$

Absolutely convergent set (ACS): $(-1,1)$ and convergent set (CS): $[-1,1)$

Because for $x = -1$: $\sum_{n=1}^\infty \frac{(-1)^n}{n + \sqrt{n}}$

Apply the Alternating Test (Leibniz Test):

Check that:
1. $\lim_{n \to \infty} \frac{1}{n + \sqrt{n}} = 0.$
2. $\frac{1}{n + \sqrt{n}}$ is decreasing.
Conclusion:

By the Leibniz test, the series is conditionally convergent.

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