Power Series

$\left|\dfrac{a_{n+1}}{a_n}\right| = \dfrac{|x - 3|}{2} \cdot \dfrac{n}{n + 1} \to \dfrac{|x - 3|}{2}$.

$\dfrac{|x - 3|}{2} < 1$ means $|x - 3| < 2$, so $R = 2$ and $1 < x < 5$.

$x = 5$: $\sum \dfrac{1}{n}$ diverges (harmonic). $x = 1$: $\sum \dfrac{(-1)^{n}}{n}$ converges (alternating series test).

Include $1$, exclude $5$.

$\left|\dfrac{a_{n+1}}{a_n}\right| = |2x - 1| \cdot \dfrac{n^{2}}{(n + 1)^{2}} \to |2x - 1|$.

$|2x - 1| < 1$ means $\left|x - \tfrac{1}{2}\right| < \tfrac{1}{2}$, so $R = \tfrac{1}{2}$ and $0 < x < 1$.

$x = 1$: $\sum \dfrac{1}{n^{2}}$ converges ($p = 2$). $x = 0$: $\sum \dfrac{(-1)^{n}}{n^{2}}$ converges absolutely.

Include both endpoints.

$\left|\dfrac{a_{n+1}}{a_n}\right| = \dfrac{x^{2}}{9}\sqrt{\dfrac{n}{n + 1}} \to \dfrac{x^{2}}{9}$.

$\dfrac{x^{2}}{9} < 1$ means $|x| < 3$, so $R = 3$.

$x = \pm 3$: $x^{2n} = 9^{n}$, giving $\sum \dfrac{1}{\sqrt{n}}$, which diverges ($p = \tfrac{1}{2}$).

Exclude both endpoints.

$\left|\dfrac{a_{n+1}}{a_n}\right| = \dfrac{|x + 4|}{(2n + 2)(2n + 1)} \to 0$ for every $x$.

$0 < 1$ always, so $R = \infty$.

No endpoints to test.

Every real $x$.

$\dfrac{x^{2}}{1 - 3x} = x^{2} \cdot \dfrac{1}{1 - 3x}$, with $r = 3x$.

$\dfrac{1}{1 - 3x} = \sum_{n=0}^{\infty} 3^{n}x^{n}$.

$x^{2}\sum_{n=0}^{\infty} 3^{n}x^{n} = \sum_{n=0}^{\infty} 3^{n}x^{n+2}$.

$|3x| < 1$, so $|x| < \tfrac{1}{3}$.

$\dfrac{1}{4 + x^{2}} = \dfrac{1}{4} \cdot \dfrac{1}{1 - \left(-\tfrac{x^{2}}{4}\right)}$.

$\dfrac{1}{4}\sum_{n=0}^{\infty} \left(-\dfrac{x^{2}}{4}\right)^{n}$.

$= \sum_{n=0}^{\infty} \dfrac{(-1)^{n}x^{2n}}{4^{n+1}}$.

$\left|\tfrac{x^{2}}{4}\right| < 1$ means $|x| < 2$.

$5 - x = 3 - (x - 2)$, so $\dfrac{1}{5 - x} = \dfrac{1}{3} \cdot \dfrac{1}{1 - \frac{x - 2}{3}}$.

$\dfrac{1}{3}\sum_{n=0}^{\infty} \left(\dfrac{x - 2}{3}\right)^{n}$.

$= \sum_{n=0}^{\infty} \dfrac{(x - 2)^{n}}{3^{n+1}}$.

$\left|\dfrac{x - 2}{3}\right| < 1$ means $-1 < x < 5$.

$\dfrac{1}{(1 + x)^{2}} = -\dfrac{d}{dx}\left(\dfrac{1}{1 + x}\right)$.

$\dfrac{1}{1 + x} = \sum_{n=0}^{\infty} (-1)^{n}x^{n}$ for $|x| < 1$.

$-\dfrac{d}{dx}\sum_{n=0}^{\infty} (-1)^{n}x^{n} = \sum_{n=1}^{\infty} (-1)^{n+1}n\,x^{n-1}$.

No constant to fix (we differentiated). Still $|x| < 1$.

$\dfrac{d}{dx}\log(1 + x^{2}) = \dfrac{2x}{1 + x^{2}}$, so $\log(1 + x^{2}) = \int \dfrac{2x}{1 + x^{2}}\,dx$.

$\dfrac{2x}{1 + x^{2}} = 2x\sum_{n=0}^{\infty} (-1)^{n}x^{2n} = \sum_{n=0}^{\infty} 2(-1)^{n}x^{2n+1}$ for $|x| < 1$.

$\int \sum_{n=0}^{\infty} 2(-1)^{n}x^{2n+1}\,dx = C + \sum_{n=0}^{\infty} \dfrac{(-1)^{n}x^{2n+2}}{n+1}$.

At $x = 0$: $\log 1 = 0 = C$.

$F$ is the integral of $\dfrac{t}{1 - t^{3}}$, a geometric-type fraction.

$\dfrac{t}{1 - t^{3}} = t\sum_{n=0}^{\infty} t^{3n} = \sum_{n=0}^{\infty} t^{3n+1}$ for $|t| < 1$.

$\int_0^{x} t^{3n+1}\,dt = \dfrac{x^{3n+2}}{3n+2}$.

The lower limit $0$ already fixes the constant: $F(0) = 0$.

$\sum_{n=1}^{\infty} n x^{n}$ at $x = \tfrac{1}{6}$.

$\sum_{n=1}^{\infty} n x^{n-1} = \dfrac{d}{dx}\sum_{n=0}^{\infty} x^{n}$: differentiate.

$\dfrac{d}{dx}\dfrac{1}{1 - x} = \dfrac{1}{(1 - x)^{2}}$, so $\sum_{n=1}^{\infty} n x^{n} = \dfrac{x}{(1 - x)^{2}}$.

$\dfrac{1/6}{(5/6)^{2}} = \dfrac{1}{6} \cdot \dfrac{36}{25} = \dfrac{6}{25}$.

$\sum_{n=2}^{\infty} n(n - 1)x^{n}$ at $x = \tfrac{1}{2}$.

$\sum_{n=2}^{\infty} n(n - 1)x^{n-2} = \dfrac{d^{2}}{dx^{2}}\sum_{n=0}^{\infty} x^{n}$: differentiate twice.

$\dfrac{d^{2}}{dx^{2}}\dfrac{1}{1 - x} = \dfrac{2}{(1 - x)^{3}}$, so $\sum_{n=2}^{\infty} n(n - 1)x^{n} = \dfrac{2x^{2}}{(1 - x)^{3}}$.

$\dfrac{2 \cdot \frac{1}{4}}{\frac{1}{8}} = 4$.

$\sum_{n=1}^{\infty} \dfrac{x^{n}}{n}$ at $x = \tfrac{1}{5}$.

$\sum_{n=1}^{\infty} \dfrac{x^{n}}{n} = \int_0^{x} \sum_{n=1}^{\infty} t^{n-1}\,dt$: integrate.

$\int_0^{x} \dfrac{dt}{1 - t} = -\log(1 - x)$, so $\sum_{n=1}^{\infty} \dfrac{x^{n}}{n} = -\log(1 - x)$.

$-\log\left(1 - \tfrac{1}{5}\right) = \log\dfrac{5}{4}$.