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Homework 10

Mos Kullathon
921425216

1. Determine whether the following series is convergent or divergent. We may use any tests available

(i) $\sum_{n=1}^\infty \frac{1}{n^2 2^n}$

As a $p$-series where $p=2$, $\displaystyle\sum_{n=1}^\infty\frac{1}{n^2}$ converges. Since $\displaystyle\frac{1}{n^2}\ge\frac{1}{n^22^n}$ for $n\ge1$, by comparison test $\displaystyle\sum_{n=1}^\infty \frac{1}{n^22^n}$ converges.

(ii) $\sum_{n=2}^\infty \frac{3n^3+2n^2+2}{n^4-1}$

Let $\displaystyle a_n=\frac{3n^3+2n^2+2}{n^4-1}$ and $\displaystyle b_n=\frac{1}{n}$.

Then,
$$\frac{a_n}{b_n} =\frac{n(3n^3+2n^2+2)}{n^4-1} =\frac{3n^4+2n^3+2n}{n^4-1} \\ \therefore \lim_{n\to\infty}\frac{a_n}{b_n} =\lim_{n\to\infty}\frac{3n^4+2n^3+2n}{n^4-1}=3.$$

Since $\displaystyle\frac{a_n}{b_n}\to 3$ and $\displaystyle\sum_{n=1}^\infty b_n = \sum_{n=1}^\infty\frac{1}{n}$ is a $p$-series such that $p=1$ (and as such, diverges), by limit comparison test $\displaystyle\sum_{n=2}^\infty \frac{3n^3+2n^2+2}{n^4-1}$ also diverges.

(iii) $\sum_{n=1}^\infty \frac{n^{1/3}}{(n^{3/2}-1)^{1/2}}$

The first term is undefined. Therefore, the sum does not exist.
$$n=1\implies\frac{1^{1/3}}{(1^{3/2}-1)^{1/2}}=\frac{1}{0}$$

(iv) $\sum_{n=1}^\infty \frac{n^2\ln n}{n^5+2n-1}$

Let $\displaystyle a_n = \frac{n^2\ln n}{n^5+2n-1}$ and $\displaystyle b_n = \frac{n^\epsilon}{n^3}$ for some small values of $\displaystyle\epsilon>0$.

Then,
$$\frac{a_n}{b_n} = \frac{n^2\ln n}{n^5+2n-1}\cdot \frac{n^3}{n^\epsilon} = \frac{n^5\ln n}{n^\epsilon(n^5+2n-1)} \\ \therefore \lim_{n\to\infty}\frac{a_n}{b_n} = \lim_{n\to\infty}\frac{\ln n}{n^\epsilon} = 0.$$

Since $\displaystyle\frac{a_n}{b_n}\to0$ and $\displaystyle\sum_{n=0}^\infty b_n = \sum_{n=0}^\infty\frac{n^\epsilon}{n^3}$ converges by $p$-series for $0<\epsilon<2$, then by limit comparison test, $\displaystyle\sum_{n=0}^\infty\frac{n^2\ln n}{n^5+2n-1}$ also converges.

(v) $\sum_{n=1}^\infty (-1)^{n^2}\frac{n+1}{n!}$

Let $\displaystyle a_n=(-1)^{n^2}\frac{n+1}{n!}$.

Then,
$$a_{n+1}=\frac{(-1)^{(n+1)^2}(n+1)+1}{(n+1)!} =\frac{(-1)^{n^2+2n+1}(n+2)}{(n+1)n!}$$

and
$$\begin{align*} \left|\frac{a_{n+1}}{a_n}\right| &= \left| \frac{(-1)^{\cancel{n^2}+2n+1}(n+2)} {(n+1)\cancel{n!}} \cdot \frac{\cancel{n!}}{\cancel{(-1)^{n^2}}(n+1)} \right| \\ &= \left| \frac{(-1)^{2n+1}(n+2)}{(n+1)^2} \right| \end{align*} \\ \therefore\lim_{n\to\infty} \left|\frac{a_{n+1}}{a_n}\right|=0<1.$$

By ratio test, the series converges absolutely.

(vi) $\sum_{n=1}^\infty \frac{(n!)^3}{(3n)!}$

Let $\displaystyle a_n=\frac{(n!)^3}{(3n)!}$.

Then,
$$\begin{align*} a_{n+1} &=\frac{((n+1)!)^3}{(3(n+1))!} \\ &=\frac{((n+1)!)^3}{(3n+3)!} \\ &= \frac{((n+1)n!)^3}{(3n+3)(3n+2)(3n+1)(3n)!} \\ &= \frac{(n+1)^3(n!)^3}{(3n+3)(3n+2)(3n+1)(3n)!} \end{align*}$$

and
$$\begin{align*} \left|\frac{a_{n+1}}{a_n}\right| &= \left| \frac{(n+1)^3\cancel{(n!)^3}} {(3n+3)(3n+2)(3n+1)\cancel{(3n)!}} \cdot \frac{\cancel{(3n)!}}{\cancel{(n!)^3}} \right| \\ &= \left| \frac{(n+1)^3} {(3n+3)(3n+2)(3n+1)} \right| \end{align*} \\ \therefore \lim_{n\to\infty}\left|\frac{a_{n+1}}{a_n}\right|=\frac{1}{3^3}=\frac{1}{27}<1.$$

By ratio test, the series converges absolutely.

(vii) $\sum_{n=1}^\infty (\frac{n}{3n+1})^n$

Let $\displaystyle a_n = \(\frac{n}{3n+1}\)^n$.

Then,
$$|a_n|^{\frac{1}{n}} =\left|\(\frac{n}{3n+1}\)^n\right|^\frac{1}{n} =\left|\frac{n}{3n+1}\right| =\frac{n}{3n+1}, \quad n\ge 1$$

and
$$\lim_{n\to\infty}|a_n|^\frac{1}{n} = \lim_{n\to\infty}\frac{n}{3n+1}=\frac{1}{3}<1.$$

By root test, the series converges absolutely.

(viii) $\sum_{n=2}^\infty \frac{(-1)^n}{\ln n}$

Let $\displaystyle a_n = \frac{1}{\ln n}$.

Since $\ln n$ is monotonically increasing, its reciprocal $\displaystyle a_n = \frac{1}{\ln n}$ must be monotonically decreasing.

Then,
$$\lim_{n\to\infty}a_n =\lim_{n\to\infty}\frac{1}{\ln n} =0.$$

By alternating series test, the series converges.

(ix) $\sum_{n=2}^\infty (-1)^n\ln n$

$$\lim_{n\to\infty}\ln n = \infty$$

By divergence test, the series diverges.

2. Find the radius of convergence and interval of convergence for the following power series

(i) $\sum_{n=0}^\infty\frac{1}{n2^n}x^n$

$$a_n=\frac{x^n}{n2^n},\quad a_{n+1}=\frac{x^{n+1}}{(n+1)2^{n+1}} \\ \begin{align*} \left| \frac{a_{n+1}}{a_n} \right| &= \left| \frac{x^{n+1}}{(n+1)2^{n+1}} \cdot\frac{n2^n}{x^n} \right| \\ &= \left| \frac{xn}{2(n+1)} \right| \\ &= |x|\frac{n}{2n+2},\quad n\ge0 \end{align*} \\ \therefore L=\lim_{n\to\infty} \left| \frac{a_{n+1}}{a_n} \right| = \frac{|x|}{2} \implies L <1 \iff |x|<2$$

Since
$$|x|<2\iff -2<x<2,$$

by ratio test, the series converges absolutely for $-2<x<2$ centered at $x=0$.

$$x=-2\implies \sum_{n=0}^\infty\frac{1}{n2^n}(-2)^n =\sum_{n=0}^\infty\frac{(-1)^n}{n}$$

Since $\displaystyle\frac{1}{n}\to0$, by alternating series test, the series converges at $x=-2$.

$$x=2\implies \sum_{n=0}^\infty\frac{1}{n2^n}(2)^n =\sum_{n=0}^\infty\frac{1}{n}$$

At $x=2$, the series is a $p$-series such that $p=1$. Therefore, the series diverges at $x=2$.

As such:

• the radius of convergence is $2$
• the interval of convergence is $[-2,2)$.

(ii) $\sum_{n=0}^\infty\frac{(x-1)^n}{\sqrt{n}}$

$$a_n=\frac{(x-1)^n}{\sqrt{n}},\quad a_{n+1}=\frac{(x-1)^{n+1}}{\sqrt{n+1}} \\ \begin{align*} \left|\frac{a_{n+1}}{a_n}\right| &= \left| \frac{(x-1)^{n+1}}{\sqrt{n+1}} \cdot\frac{\sqrt{n}}{(x-1)^n} \right| \\ &= \left| \frac{\sqrt{n}(x-1)}{\sqrt{n+1}} \right| \\ &= |x-1|\frac{\sqrt{n}}{\sqrt{n+1}},\quad n\ge0 \end{align*} \\ \therefore L=\lim_{n\to\infty} \left|\frac{a_{n+1}}{a_n}\right|=|x-1| \implies L<1 \iff |x-1|<1 \\$$

Since
$$|x-1|<1 \iff -1<x-1<1 \iff 0<x<2,$$

by ratio test, the series converges absolutely for $0<x<2$ centered at $x=1$.

$$x=0\implies\sum_{n=0}^\infty\frac{(0-1)^n}{\sqrt{n}} =\sum_{n=0}^\infty\frac{(-1)^n}{\sqrt{n}}$$

Since $\displaystyle\frac{1}{\sqrt{n}}\to0$, by alternating series test, the series converges at $x=0$.

$$x=2\implies\sum_{n=0}^\infty\frac{(2-1)^n}{\sqrt{n}} =\sum_{n=0}^\infty\frac{1^n}{\sqrt{n}} =\sum_{n=0}^\infty\frac{1}{n^{1/2}}$$

At $x=2$, the series is a $p$-series such that $\displaystyle p=\frac{1}{2}$. Therefore, the series diverges at $x=2$.

As such:

• the radius of convergence is $1$
• the interval of convergence is $[0,2)$.

(iii) $\sum_{n=2}^\infty \frac{1}{n\ln n}x^n$

$$a_n=\frac{x^n}{n\ln n},\quad a_{n+1}=\frac{x^{n+1}}{(n+1)\ln(n+1)} \\ \begin{align*} \left|\frac{a_{n+1}}{a_n}\right| &= \left| \frac{x^{n+1}}{(n+1)\ln(n+1)}\cdot \frac{n\ln n}{x^n} \right| \\ &= \left| \frac{nx\ln n}{(n+1)\ln(n+1)} \right| \end{align*} \\ \begin{align*} \therefore L = \lim_{n\to\infty}\left|\frac{a_{n+1}}{a_n}\right| &= \lim_{n\to\infty}\left| \frac{nx\ln n}{(n+1)\ln(n+1)} \right| \\ &= \lim_{n\to\infty} \left| \frac{nx}{n+1} \right| \cdot\lim_{n\to\infty} \left|\frac{\ln n}{\ln(n+1)}\right| \\ &= |x| \cdot \lim_{n\to\infty} \left|\frac{\frac{1}{n}}{\frac{1}{n+1}}\right| \\ &= |x| \\ \implies L <1 &\iff |x| <1 \end{align*}$$

Since
$$|x|<1\iff-1<x<1,$$

by ratio test, the series converges absolutely for $-1<x<1$.

$$x=-1\implies \sum_{n=2}^\infty \frac{1}{n\ln n}(-1)^n$$

Since $\displaystyle\frac{1}{n\ln n}\to0$, by alternating series test, the series converges at $x=-1$.

$$x=1\implies \sum_{n=2}^\infty \frac{1}{n\ln n}(1)^n =\sum_{n=2}^\infty \frac{1}{n\ln n}$$

Let $\displaystyle g(x)=\frac{1}{x\ln x}$. Since $g'(x)<0$ for $x\ge2$, we apply the integral test.

$$\begin{darray}{cc} u=\ln x &\implies& \dd u=\frac{1}{x}\dd x \\ &\iff&\dd x = x\dd u \\ x=2 &\implies& u = \ln 2 \\ x=n &\implies& u = \ln n \end{darray} \\ \begin{align*} \lim_{n\to\infty}\int_2^n\frac{1}{x\ln x}\dd x &= \lim_{n\to\infty}\int_{\ln 2}^{\ln n} \frac{x\dd u}{xu} \\ &= \lim_{n\to\infty} \bigg[\ln u\bigg]_{\ln 2}^{\ln n} \\ &= \lim_{n\to\infty}\ln(\ln n)-\ln(\ln 2) \\ &= \infty \end{align*}$$

By integral test, the series diverges at $x=1$.

As such:

• the radius of convergence is $1$
• the interval of convergence is $[-1,1)$.

(iv) $\sum_{n=2}^\infty (2+\frac{1}{n})^n x^n$

$$a_n = \(2+\frac{1}{n}\)^n x^n \\ \begin{align*} |a_n|^\frac{1}{n} &= \left| \(2+\frac{1}{n}\) x \right| \end{align*} \\ \therefore L=\lim_{n\to\infty}|a_n|^\frac{1}{n} = |2x| \implies L < 1 \iff |2x|<1$$

Since
$$|2x|<1 \iff |x| < \frac{1}{2} \iff -\frac{1}{2}<x<\frac{1}{2},$$

by root test, the series converges absolutely for $\displaystyle-\frac{1}{2}<x<\frac{1}{2}$ centered at $x=0$.

$$x=-\frac{1}{2}\implies \sum_{n=2}^\infty \(2+\frac{1}{n}\)^n \(-\frac{1}{2}\)^n \\ a_n=\(2+\frac{1}{n}\)^n \(-\frac{1}{2}\)^n \\ |a_n|^\frac{1}{n} = \left| \(2+\frac{1}{n}\) \(-\frac{1}{2}\) \right|$$

Since $|a_n|^\frac{1}{n}\to-1$, by root test, the series converges at $\displaystyle x=-\frac{1}{2}$.

$$x=\frac{1}{2}\implies \sum_{n=2}^\infty \(2+\frac{1}{n}\)^n \(\frac{1}{2}\)^n \\ \begin{align*} \lim_{n\to\infty}\(2+\frac{1}{n}\)^n \(\frac{1}{2}\)^n &= \lim_{n\to\infty} \(\(2+\frac{1}{n}\) \(\frac{1}{2}\)\)^n \\ &= \lim_{n\to\infty} \(1+\frac{1}{2n}\)^n \\ &= \lim_{n\to\infty} e^{n\ln\(1+\frac{1}{2n}\) } \\ &= \lim_{n\to\infty} \exp\frac{\ln\(1+\frac{1}{2n}\) }{1/n} \\ &= \exp\lim_{n\to\infty} \frac{\frac{1}{1+\frac{1}{2n}}\cdot -\frac{1}{2n^2}}{-\frac{1}{n^2}} \\ &= \exp\lim_{n\to\infty} \frac{1}{1+\frac{1}{2n}}\cdot\frac{1}{2} \\ &= \exp\frac{1}{2} \\ &= \sqrt{e} \quad\approx 1.64872 \end{align*}$$

Since $\displaystyle\(2+\frac{1}{n}\)^n \(\frac{1}{2}\)^n\to\sqrt{e}$, by divergence test, the series diverges at $\displaystyle x=\frac{1}{2}$.

As such:

• the radius of convergence is $\displaystyle\frac{1}{2}$
• the interval of convergence is $\displaystyle\[-\frac{1}{2},\frac{1}{2}\)$.

3. Section 6.1

In the following exercises, given that $\displaystyle\frac{1}{1-x}=\sum_{n=0}^\infty x^n$ with convergence in $(−1,1)$, find the power series for each function with the given center a, and identify its interval of convergence.

(35) $f(x)=\frac{x}{1-x^2}; a=0$

$$\begin{align*} f(x) &= \frac{x}{1-x^2} \\ &= \sum_{n=0}^\infty x(x^2)^n,\quad |x^2|<1 \\ &= \sum_{n=0}^\infty x^{2n+1},\quad |x|<1 \end{align*}$$

Since
$$|x|<1\iff-1<x<1,$$

$f(x)$ converges within $-1<x<1$.

Then,
$$f(-1) = \sum_{n=0}^\infty(-1)^{2n+1}$$

Since $\displaystyle\lim_{n\to\infty} (-1)^{2n+1}$ does not exist, $f(-1)$ diverges by divergence test.

$$f(1) = \sum_{n=0}^\infty1^{2n+1} \\$$

Since $\displaystyle\lim_{n\to\infty}1^{2n+1}=1$, $f(1)$ diverges by divergence test.

As such,
$$f(x) =\frac{x}{1-x^2} =\sum_{n=0}^\infty x^{2n+1},\quad x\in(-1,1).$$

(37) $f(x)=\frac{x^2}{1+x^2}; a=0$

$$\begin{align*} f(x) &= \frac{x^2}{1+x^2} \\ &= \frac{x^2}{1-(-x^2)} \\ &= \sum_{n=0}^\infty x^2(-x^2)^n, \quad\left|-x^2\right|<1 \\ &= \sum_{n=0}^\infty(-1)^n x^{2n+2}, \quad|x|<1 \end{align*}$$

Since
$$|x|<1\iff-1<x<1,$$

$f(x)$ converges within $-1<x<1$.

Then,
$$f(-1)=\sum_{n=0}^\infty(-1)^n (-1)^{2n+2} =\sum_{n=0}^\infty(-1)^{3n+2}$$

Since $\displaystyle\lim_{n\to\infty}(-1)^{3n+2}$ does not exist, $f(-1)$ diverges by divergence test.

$$f(1)=\sum_{n=0}^\infty(-1)^n (1)^{2n+2} =\sum_{n=0}^\infty(-1)^n$$

Since $\displaystyle\lim_{n\to\infty}(-1)^n$ does not exist, $f(1)$ diverges by divergence test.

As such,
$$f(x)=\frac{x^2}{1+x^2} =\sum_{n=0}^\infty(-1)^n x^{2n+2}, \quad x\in(-1,1)$$

(39) $f(x)=\frac{1}{1-2x}; a=0$

$$\begin{align*} f(x) &= \frac{1}{1-2x} \\ &= \sum_{n=0}^\infty(2x)^n, \quad\left|2x\right|<1 \\ &= \sum_{n=0}^\infty2^n x^n, \quad|x|<\frac{1}{2} \end{align*}$$

Since
$$|x|<\frac{1}{2}\iff-\frac{1}{2}<x<\frac{1}{2},$$

$f(x)$ converges within $\displaystyle-\frac{1}{2}<x<\frac{1}{2}$.

Then,
$$f\(-\frac{1}{2}\) =\sum_{n=0}^\infty 2^n\(-\frac{1}{2}\)^n =\sum_{n=0}^\infty (-1)^n$$

Since $\displaystyle\lim_{n\to\infty}(-1)^n$ does not exist, $f\(-\frac{1}{2}\)$ diverges by divergence test.

$$f\(\frac{1}{2}\) =\sum_{n=0}^\infty2^n\(\frac{1}{2}\)^n =\sum_{n=0}^\infty 1$$

Since $\displaystyle\lim_{n\to\infty}1=1$, $f(1)$ diverges by divergence test.

As such,
$$f(x)=\frac{1}{1-2x}=\sum_{n=0}^\infty2^n x^n, \quad x\in\(-\frac{1}{2},\frac{1}{2}\)$$

(41) $f(x)=\frac{x^2}{1-4x^2}; a=0$

$$\begin{align*} f(x) &= \frac{x^2}{1-4x^2} \\ &= \sum_{n=0}^\infty x^2(4x^2)^n,\quad |4x^2|<1 \\ &= \sum_{n=0}^\infty x^24^nx^{2n},\quad |x^2|<\frac{1}{4} \\ &= \sum_{n=0}^\infty 4^nx^{2n+2},\quad |x|<\frac{1}{2} \end{align*}$$

Since
$$|x|<\frac{1}{2}\iff-\frac{1}{2}<x<\frac{1}{2},$$

$f(x)$ converges within $\displaystyle-\frac{1}{2}<x<\frac{1}{2}$.

Then,
$$f\(-\frac{1}{2}\)=\sum_{n=0}^\infty 4^n \(-\frac{1}{2}\)^{2n+2}.$$

Since $\displaystyle\lim_{n\to\infty}4^n\(-\frac{1}{2}\)^{2n+2}$ does not exist, by divergence test $\displaystyle f\(-\frac{1}{2}\)$ diverges.

$$f\(\frac{1}{2}\)=\sum_{n=0}^\infty 4^n \(\frac{1}{2}\)^{2n+2}$$

Since
$$\begin{align*} \lim_{n\to\infty}4^n\(\frac{1}{2}\)^{2n+2} &= \lim_{n\to\infty}\frac{4^n}{2^{2n+2}} \\ &= \lim_{n\to\infty}\frac{4^n}{2^{2n}\cdot2^2} \\ &= \lim_{n\to\infty}\frac{4^n}{(2^2)^n\cdot2^2} \\ &= \lim_{n\to\infty}\frac{1}{4} \\ &= \frac{1}{4} \end{align*}$$

by divergence test $\displaystyle f\(\frac{1}{2}\)$ diverges.

As such,
$$f(x)=\frac{x^2}{1-4x^2} =\sum_{n=0}^\infty 4^nx^{2n+2}, \quad x\in\(-\frac{1}{2},\frac{1}{2}\)$$

4. Find

(i) The Taylor polynomial of degree 3 of $f(x)=x^{2/3}$ at $x=8$

$$\begin{align*} f(x)=x^{2/3} &\implies f'(x)=\frac{2}{3}x^{-1/3} \\ &\implies f''(x)=-\frac{2}{9}x^{-4/3} \\ &\implies f'''(x)=\frac{8}{27}x^{-7/3} \\ \end{align*} \\ \begin{darray}{cll} f(8) &= 8^{2/3} &= 4\\ f'(8) &= \frac{2}{3}\cdot 8^{-1/3} &= \frac{1}{3}\\ f''(8) &= -\frac{2}{9}\cdot 8^{-4/3} &= -\frac{1}{72}\\ f'''(8) &= \frac{8}{27}\cdot 8^{-7/3} &= \frac{1}{432}\\ \end{darray} \\ \begin{align*} \therefore\sum_{n=0}^3\frac{f^{(n)}(x)}{n!}(x-8)^n &= f(8) + f'(8)(x-8) +\frac{f''(8)(x-8)^2}{2!} +\frac{f''(8)(x-8)^3}{3!} \\ &= 4 +\frac{x-8}{3} -\frac{(x-8)^2}{72\cdot2!} +\frac{(x-8)^3}{432\cdot3!} \\ &= 4 +\frac{x-8}{3} -\frac{(x-8)^2}{144} +\frac{(x-8)^3}{2592} \end{align*}$$

(ii) The Taylor polynomial of degree 2 of $f(x)=\sec x$ at $x=0$

$$\begin{align*} f(x)=\sec x &\implies f'(x) = \sec x\tan x \\ &\implies f''(x) = (\sec x\tan^2 x) + \sec^3x \\ \end{align*} \\ \begin{darray}{cll} f(0) &= \sec 0 &= 1\\ f'(0) &= \sec 0\tan 0 &= 0\\ f''(0) &= (\sec 0\tan^2 0) + \sec^30 &= 1\\ \end{darray} \\ \begin{align*} \therefore\sum_{n=0}^2\frac{f^{(n)}(x)}{n!}(x-0)^n &= f(0) + f'(0)x +\frac{f''(0)x^2}{2!} \\ &= 1 + 0x + \frac{x^2}{2!} \\ &= 1 + \frac{x^2}{2} \end{align*}$$

5. Find the Taylor series expansion of the following functions using the formula $$\frac{1}{1-x}=\sum_{n=0}^\infty x^n$$

Identify their interval of convergence. You may need to use differentiation and integration of the Taylor series.

Center of convergence not specified
The following answers for questions 5(i) through 5(iii) assumes that the center of convergence is at $x=0$.

(i) $\frac{1}{2+3x}$

Let the center of convergence be at $x=0$.

$$\begin{align*} \frac{1}{2+3x} &= \frac{1}{2(1+\frac{3}{2}x)} \\ &= \frac{1}{2}\cdot\frac{1}{1-(-\frac{3}{2}x)} \\ &= \sum_{n=0}^\infty\frac{1}{2}\(-\frac{3}{2}x\)^n, \quad\left|-\frac{3}{2}x\right| < 1 \\ &= \sum_{n=0}^\infty \frac{(-3)^n}{2(2)^n} x^n,\quad|x|<\frac{2}{3} \\ &= \sum_{n=0}^\infty \frac{(-3)^n}{2^{n+1}} x^n,\quad|x|<\frac{2}{3} \end{align*}$$

Since
$$|x|<\frac{2}{3}\iff-\frac{2}{3}<x<\frac{2}{3},$$
the series converges within $\displaystyle-\frac{2}{3}<x<\frac{2}{3}$.

Then,
$$x=-\frac{2}{3} \implies\sum_{n=0}^\infty\frac{(-3)^n}{2^{n+1}} \(-\frac{2}{3}\)^n = \sum_{n=0}^\infty\frac{2^n}{2^{n+1}}$$

Since $\displaystyle\frac{2^n}{2^{n+1}}\to\frac{1}{2}$, the series diverges by divergence test at $\displaystyle x=-\frac{2}{3}$.

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

Since $\displaystyle\frac{2^n}{2^{n+1}}\to\frac{1}{2}$, the series diverges by divergence (and therefore the limit of the summand does not exist) test at $\displaystyle x=\frac{2}{3}$.

As such, the interval of convergence for a series centered at $x=0$ is $\displaystyle\(-\frac{2}{3},\frac{2}{3}\)$.

(ii) $\frac{1}{(1-2x)^2}$

Let the center of convergence be at $x=0$.

$$\begin{align*} \frac{1}{1-2x} &= \sum_{n=0}^\infty(2x)^n, \quad|2x|<1 \\ &= \sum_{n=0}^\infty 2^n x^n, \quad|x|<\frac{1}{2} \\ \end{align*}$$

Since
$$\frac{\dd}{\dd x}\(\frac{1}{1-2x}\) = \frac{1}{(1-2x)^2},$$

then
$$\begin{align*} \frac{1}{(1-2x)^2} &= \sum_{n=1}^\infty 2n(2x)^{n-1},\quad |x|<\frac{1}{2} \\ &= \sum_{n=1}^\infty 2^nn x^{n-1},\quad |x|<\frac{1}{2}. \end{align*}$$

And since
$$|x|<-\frac{1}{2}\iff-\frac{1}{2}<x<\frac{1}{2},$$

the series converges within $\displaystyle-\frac{1}{2}<x<\frac{1}{2}$.

Then,
$$x=-\frac{1}{2}\implies \sum_{n=1}^\infty(-1)^{n-1}2n$$

Since $2n\to\infty$, the series diverges by divergence test at $\displaystyle x=-\frac{1}{2}$.

$$x=\frac{1}{2}\implies \sum_{n=1}^\infty 2n$$

Since $2n\to\infty$, the series diverges by divergence test at $\displaystyle x=\frac{1}{2}$.

As such, the interval of convergence for a series centered at $x=0$ is $\displaystyle\(-\frac{1}{2},\frac{1}{2}\)$.

(iii) $\ln(1+x)$

Let the center of convergence be at $x=0$.

$$\frac{1}{1+x} = \sum_{n=0}^\infty(-1)^n x^n,\quad |x|<1$$

Since
$$\int\frac{1}{1+x}\dd x = \ln(1+x),$$

then
$$\begin{align*} \ln(1+x) &= \sum_{n=0}^\infty\frac{(-1)^nx^{n+1}}{n+1}, \quad |x|<1. \end{align*}$$

And since
$$|x|<1\iff-1<x<1,$$

the series converges within $-1<x<1$.

Then,
$$x=-1\implies \sum_{n=0}^\infty(-1)^{2n+1}\frac{1}{n+1} =\sum_{n=0}^\infty -\frac{1}{n+1}$$

Since $\displaystyle-\frac{1}{n+1}\approx-\frac{1}{n}$ for large values of $n$, the series diverges at $x=-1$ by $p$-series test.

$$x=1\implies \sum_{n=0}^\infty(-1)^n\frac{1}{n+1}$$

Since $\displaystyle\frac{1}{n+1}\to0$, the series converge at $x=1$ by alternating series test.

As such, the interval of convergence for a series centered at $x=0$ is $(-1,1]$.