The ratio test for series

## Power series
            ---
            ## Power series
            #### Definition
            If $a\in\mathbb R$ and $(c_n)\_{n=0}^\infty$ is a sequence of real numbers, then the function of $x$ given by
            $\displaystyle \sum\_{n=0}^\infty c_n (x-a)^n$
            is a *power series centred at* $a$.
            Often $a=0$.
            
            ---
            ## Power series
            #### Definition
            If $a\in\mathbb R$ and $(c_n)\_{n=0}^\infty$ is a sequence of real numbers, then the function of $x$ given by
            $\displaystyle \sum\_{n=0}^\infty c_n (x-a)^n$
            is a *power series centred at* $a$.
            Often $a=0$.

The power series may converge for some $x$ and diverge for others.
            #### Example
            $\displaystyle \sum\_{n=0}^\infty \frac{x^n}{3^n}$ converges for $\lvert x \rvert \lt 3$ and diverges for $\lvert x \rvert \geqslant 3$.
            
            ---
            ## Power series
            #### Definition
            If $a\in\mathbb R$ and $(c_n)\_{n=0}^\infty$ is a sequence of real numbers, then the function of $x$ given by
            $\displaystyle \sum\_{n=0}^\infty c_n (x-a)^n$
            is a *power series centred at* $a$.
            Often $a=0$.

Proof:
            $\displaystyle \left\lvert\frac{x^{n+1}/3^{n+1}}{x^n/3^n}\right\rvert = \frac{\lvert x\rvert}3 \underset{n\to\infty}\longrightarrow \frac{\lvert x\rvert}3$.
            Hence, by the ratio test, the power series converges for $\lvert x \rvert \lt 3$ and diverges for $\lvert x \rvert \gt 3$.
            If $\lvert x \rvert = 3$, then the power series is $1+1+1+\ldots$ or $1-1+1-\ldots$, both of which diverge.
            
            ---
            ## Power series
            #### Definition
            If $a\in\mathbb R$ and $(c_n)\_{n=0}^\infty$ is a sequence of real numbers, then the function of $x$ given by
            $\displaystyle \sum\_{n=0}^\infty c_n (x-a)^n$
            is a *power series centred at* $a$.
            Often $a=0$.

The power series may converge for some $x$ and diverge for others.

The ratio test is good for telling us whether a power series converges.
            
            ---
            ## Power series
            #### Definition
            If $a\in\mathbb R$ and $(c_n)\_{n=0}^\infty$ is a sequence of real numbers, then the function of $x$ given by
            $\displaystyle \sum\_{n=0}^\infty c_n (x-a)^n$
            is a *power series centred at* $a$.
            Often $a=0$.

The power series may converge for some $x$ and diverge for others.

The ratio test is good for telling us whether a power series converges.
            #### Theorem
            For any power series centred at $a$, there exists a unique $R \geqslant 0$ (possibly $R=\infty$) such that the power series converges for $x\in(a-R,a+R)$ and diverges for $\lvert x-a \rvert\gt R$.
            The $R$ is called the *radius of convergence* of the power series.
            #### Proof
            Do the ratio test.
            
            ---
            ## Power series
            #### Definition
            If $a\in\mathbb R$ and $(c_n)\_{n=0}^\infty$ is a sequence of real numbers, then the function of $x$ given by
            $\displaystyle \sum\_{n=0}^\infty c_n (x-a)^n$
            is a *power series centred at* $a$.
            Often $a=0$.

The power series may converge for some $x$ and diverge for others.

The ratio test is good for telling us whether a power series converges.
            #### Theorem
            For any power series centred at $a$, there exists a unique $R \geqslant 0$ (possibly $R=\infty$) such that the power series converges for $x\in(a-R,a+R)$ and diverges for $\lvert x-a \rvert\gt R$.
            The $R$ is called the *radius of convergence* of the power series.
            #### Limitation of above theorem
            This does not decide convergence at $x=a\pm R$.
            Often difficult. The ratio test will not help.
            
            ---
            ## Power series
            #### Theorem
            For any power series centred at $a$, there exists a unique $R \geqslant 0$ (possibly $R=\infty$) such that the power series converges for $x\in(a-R,a+R)$ and diverges for $\lvert x-a \rvert\gt R$.
            The $R$ is called the *radius of convergence* of the power series.
            #### Example
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Find the interval on which $\displaystyle \sum\_{n=0}^\infty \frac{x^n}{3^nn}$ converges.

Find the interval on which $\displaystyle \sum\_{n=0}^\infty \frac{(x-5)^n}{3^nn^2}$ converges.

Find the interval on which $\displaystyle \sum\_{n=0}^\infty \frac{n(x+1)^n}{3^n}$ converges.

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            <div style="flex:flex:1;margin-left:2vw">

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            ---
            ## Power series
            #### Theorem
            For any power series centred at $a$, there exists a unique $R \geqslant 0$ (possibly $R=\infty$) such that the power series converges for $x\in(a-R,a+R)$ and diverges for $\lvert x-a \rvert>R$.
            The $R$ is called the *radius of convergence* of the power series.
            #### Example
            <div style="display:flex;flex-direction:row;width:100%">
            <div style="flex:0 0 auto;margin-right:2vw;">

Find the interval on which $\displaystyle \sum\_{n=0}^\infty \frac{x^n}{3^nn}$ converges.

Find the interval on which $\displaystyle \sum\_{n=0}^\infty \frac{(x-5)^n}{3^nn^2}$ converges.

Find the interval on which $\displaystyle \sum\_{n=0}^\infty \frac{n(x+1)^n}{3^n}$ converges.

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            <div style="flex:flex:1;margin-left:2vw">

Answer: $\displaystyle [-3,3) \vphantom{\sum\_{n=0}^\infty \frac{x^n}{3^nn}}$.

Answer: $\displaystyle [2,8] \vphantom{\sum\_{n=0}^\infty \frac{(x-5)^n}{3^nn^2}}$.

Answer: $\displaystyle (-4,2) \vphantom{\sum\_{n=0}^\infty \frac{n(x+1)^n}{3^n}}$.

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