Preliminary
Definition. Given a power series $\sum_{n=0}^\infty a_nx^n$, the value $R$ is said to be the radius of convergence of this power series ifBelow is a standard result for power series!
(i) When $|x|<R$, the power series $\sum_{n=0}^\infty a_nx^n$ converges.
(ii) When $|x|>R$, the power series $\sum_{n=0}^\infty a_nx^n$ diverges.
Note that given we have found the radius of convergence $R$ of $\sum_{n=0}^\infty a_nx^n$, then the power series is known to be convergent on $(-R,R)$, but the convergence of $\sum_{n=0}^\infty a_n(-R)^n$ and $\sum_{n=0}^\infty a_nR^n$ is still uncertain, in this course we need to find the following:Result. Consider the power series $\sumn a_nx^n$. If one of the following limits \[
\ell = \limn \frac{|a_{n+1}|}{|a_n|}\quad\text{or}\quad \ell=\limn \sqrt[n]{|a_n|}
\] exists, then the following quantity\[
R=\frac{1}{\ell}
\] is the radius of convergence of $\sumn a_nx^n$.
Lastly, note that if we are given the power series $\sum_{n=0}^\infty a_n(x-a)^n$, where $a\neq 0$, then we can still compute the radius of convergence, say $R$, and then the open interval on which the power series must converge will be \[Definition. The interval of convergence of $\sumn a_nx^n$ is the set $I\subseteq \R$ on which $\sumn a_nx^n$ converges.
(a-R,a+R).
\] To see this, try to consider $x-a$ as a whole a single number, call this $u=x-a$, then \[\sum_{n=0}^\infty a_n(x-a)^n=\sum_{n=0}^\infty a_n u^n.\] We then compute $R$ as before, then the power series $\sum_{n=0}^\infty a_n u^n$ converges when $|u|<R$ and diverges when $|u|>R$, i.e.,
- $\sum_{n=0}^\infty a_n(x-a)^n$ converges when $|x-a|<R$ (i.e., $x\in (a-R,a+R)$)
- $\sum_{n=0}^\infty a_n(x-a)^n$ diverges when $|x-a|>R$ (i.e., $x\in (-\infty,a-R)\cup (a+R,\infty)$).
Problems in Tutorial Notes
Problem 1. (a) Since \[
\limn \sqrt[n]{\abs{(-1)^n\frac{3^n}{n}}}=\limn \frac{3}{\sqrt[n]{n}}=3,
\] the radius of convergence is $\boxed{R= \frac{1}{3}}$, namely, it $\sum_{n=1}^\infty (-1)^n\frac{3^n}{n}x^n$ converges on $(-R,R)$. Since $\sum_{n=1}^\infty (-1)^n\frac{3^n}{n}x^n$ diverges on $(-\infty,R)$ and $(R,\infty)$ (by part (i) of property above). To find the interval of convergence, it is enough to check whether the power series converges at $-R$ and $R$.
When $x=\frac{1}{3}$, the power series becomes $\sum_{n=1}^\infty (-1)^n \frac{1}{n}$, which converges by alternating series test. When $x=-\frac{1}{3}$, the series becomes $\sum_{n=1}^\infty \frac{1}{n}$ which is divergent. Hence the interval of convergence is $\boxed{(-\frac{1}{3},\frac{1}{3}]}$.
(b) Since \[
\limn \sqrt[n]{\frac{1}{2^{n^2}}}=\limn \frac{1}{2^n}=0,
\] the radius of convergence is $1/0=\boxed{\infty}$, i.e., the power series converges on $\R$, so $\boxed{\R}$ is also the interval of convergence.
(c) Since \[
\limn \sqrt[n]{n^n}=n=\infty,
\] the radius of convergence is $1/\infty=\boxed{0}$, i.e., the power series converges only when $x=0$. Thus the interval of convergence is $\boxed{\{0\}}$.
(d) Since \[
\limn \frac{\abs{\frac{2^{n+1}+(-1)^{n+1}}{(n+1)^2}}}{\abs{\frac{2^n+(-1)^n}{n^2}}}=\limn \abs{\frac{2+\frac{(-1)^n}{2^n}}{1+\frac{(-1)^n}{2^n}}}\frac{n^2}{(n+1)^2}=2,
\] the radius of convergence is $\boxed{\frac{1}{2}}$, i.e., the power series converges on \[\boxed{(3-\frac{1}{2},3+\frac{1}{2})}=\boxed{(2.5,3.5)}.\] When $x=3-\frac{1}{2}$, the power series becomes \[
\sum_{n=1}^\infty \brac{\frac{(-1)^n}{n^2}+\frac{1}{n^22^n}},
\] which converges. When $x=3+\frac{1}{2}$, then the series becomes \[
\sum_{n=1}^\infty \brac{\frac{1}{n^2} +\frac{(-1)^n}{n^22^n}},
\] which also converges. Thus the interval of convergence is $\boxed{[2.5,3.5]}$.
(e) When $a=0$, everything is obvious (i.e., the radius of convergence is $\infty$ and the interval of convergence is $\R$). Let's suppose $a\neq 0$, here we need to treat the power series as a function of $x^2$: \[
\sum_{n=1}^\infty \frac{a^n}{n+1}(x^2)^n.
\]
Since \[
\limn \frac{\abs{\frac{a^{n+1}}{n+2}}}{\abs{\frac{a^n}{n+1}}}=\limn |a|\frac{n+1}{n+2}=|a|,
\] the temporary radius of convergence (as a function of $x^2$) is $\boxed{\frac{1}{|a|}}$. Thus we are certain for the following:
- The power series converges when $|x|^2<\frac{1}{|a|}\iff |x|<\frac{1}{\sqrt{|a|}}$.
- The power series diverges when $|x|^2>\frac{1}{|a|}\iff |x|>\frac{1}{\sqrt{|a|}}$.
When $x=-\frac{1}{\sqrt{|a|}}$ or $x=\frac{1}{\sqrt{|a|}}$, the power series becomes \[
\sum_{n=1}^\infty \frac{\brac{\frac{a}{|a|}}^n}{n+1}=\begin{cases}
\sum_{n=1}^\infty \frac{1}{n+1},&a>0,\\
\sum_{n=1}^\infty (-1)^n\frac{1}{n+1},&a<0.
\end{cases}
\] So when $a>0$, the power series diverges at the end points. When $a<0$, the power series converges at the end points. We conclude the interval of convergence is
(i) $\dis\left(-\frac{1}{\sqrt{|a|}}, \frac{1}{\sqrt{|a|}}\right)$ when $a>0$;
(ii) $\dis\left[-\frac{1}{\sqrt{|a|}}, \frac{1}{\sqrt{|a|}}\right]$ when $a<0$.
(f) Since \[
\limn \frac{\frac{(n+1)!}{(n+1)^{n+1}}}{\frac{n!}{n^n}}=\limn \frac{1}{\brac{1+\frac{1}{n}}^n}=\frac{1}{e},
\] the radius of convergence is $\frac{1}{\frac{1}{e}}=\boxed{e}$, thus the power series converges on $(-e,e)$. When $x=-e$ and $e$, the series diverges by Stirling formula: \[
\limn \frac{n!}{\sqrt{2\pi} n^{n+1/2}e^{-n}}=1.
\]We don't go into detail. Thus the interval of convergence is $(-e,e)$.
Problem 2. We need to memorize \[
\ln(1+x)=\sum_{k=1}^\infty (-1)^{k+1} \frac{x^k}{k},
\] hence symbolically (the precise language in mathematics is formally) the power series of $g$ is \begin{align*}
g(x)&=\frac{1}{2}\ln (4-x^2)\\
&=\frac{1}{2}\brac{\ln 4+\ln \left[1-\brac{\frac{x}{2}}^2\right]}\\
&=\ln 2 +\frac{1}{2}\ln \left[1+\left(-\brac{\frac{x}{2}}^2\right)\right]\\
&=\ln 2 +\frac{1}{2}\sum_{k=1}^\infty (-1)^{k+1} \frac{1}{k}\brac{-\brac{\frac{x}{2}}^2}^k\\
&=\ln 2-\frac{1}{2}\sum_{k=1}^\infty \frac{1}{k4^k}(x^2)^k.
\end{align*} To compute the radius of convergence of this power series, the treatment below is similar to part (e) of problem 1. Let's consider $g$ as a function of $x^2$, then since \[
\limn \sqrt[n]{\frac{1}{n4^n}}=\frac{1}{4},
\] so we conclude the following: The temporary radius of convergence is $1/(1/4)=4$, so:
- The power series converges when $|x^2|<4\iff |x|<2$.
- The power series diverges when $|x^2|>4\iff |x|>2$.
Problem 3. We mainly use the formula: For every $a\in (-1,1)$, \[
\frac{1}{1-a}=1+a+a^2+\cdots.
\] Thus \begin{align*}
f&=\frac{1}{x+1}-\frac{1}{x+2}\\
&=\frac{1}{2+(x-1)}-\frac{1}{3+(x-1)}\\
&=\frac{1}{2}\frac{1}{1-(-\frac{x-1}{2})}-\frac{1}{3}\frac{1}{1-(-\frac{x-1}{3})}\\
&=\frac{1}{2}\sumn \brac{-\frac{x-1}{2}}^n-\frac{1}{3}\sumn \brac{-\frac{x-1}{3}}^n\\
&=\sumn (-1)^n \brac{\frac{1}{2^{n+1}} -\frac{1}{3^{n+1}} }(x-1)^n.
\end{align*}To compute the radius of convergence, since \[
\limn \abs{\frac{(-1)^{n+1} \brac{\frac{1}{2^{n+2}} -\frac{1}{3^{n+2}} }}{(-1)^n \brac{\frac{1}{2^{n+1}} -\frac{1}{3^{n+1}} }}}=\limn \frac{\frac{1}{2} -\brac{\frac{2}{3}}^n\frac{1}{3}}{1-\brac{\frac{2}{3}}^n}=\frac{\frac{1}{2}-0}{1-0}=\frac{1}{2},
\] so the radius of convergence is $2$.
Problem 4. The series can be written as \[
\sum_{n=2}^\infty n(n-1) \brac{-\frac{1}{2}}^n+\sumn \brac{-\frac{1}{2}}^n.
\] Consider the following power series on $(-1,1)$: \[
f(x)=\sum_{n=1}^\infty x^n=\frac{x}{1-x}.
\] It is known that when the power series converges on an open interval, it is infinitely differentiable over that open interval, hence we have \[
f''(x)=\sum_{n=2}^\infty n(n-1)x^{n-2} =\frac{2}{(1-x)^3},
\] so \[
\sum_{n=2}^\infty n(n-1)x^n=\frac{2x^2}{(1-x)^3}
\]
Put $x=-\frac{1}{2}$, \[
\sum_{n=2}^\infty n(n-1) \frac{(-1)^n}{2^n}=\frac{2(-\frac{1}{2})^2}{(1+\frac{1}{2})^3}=\frac{4}{27}.
\] Thus the answer is $\dis \frac{4}{27}+\frac{2}{3}=\frac{4+18}{27}=\frac{22}{27}$.
Recall by Taylor series expansion: \[
f(x)=\sumn \frac{f^{(n)}(a)}{n!}(x-a)^n.
\] Thus if we know the power series of $f(x)=\sumn a_n(x-a)^n$ (which happens when our power series can be obtained by other known power series expansion), then since power series expression must be unique, we have \[
\frac{f^{(n)}(a)}{n!}=a_n,\text{ i.e., } \quad f^{(n)}(a)=n!a_n.
\]
Problem 5. Since \begin{align*}
f(x)&=\frac{1}{1+(x-1)^2}\\
&=\frac{1}{1-(-(x-1)^2)}\\
&=\sumn (-(x-1)^2)^n\\
&=\sumn (-1)^n (x-1)^{2n}+\sum_{n=1}^\infty 0\cdot (x-1)^{2n-1}.
\end{align*} So if we compare coefficients, we have:
If $n$ is even, $f^{(n)}(1)=(-1)^{n/2} n!$.
If $n$ is odd, $f^{(n)}(1)=0$.
Problem 6. Integrate both sides of \[
\frac{1}{1+y^2}=\frac{1}{1-(-y^2)}=\sumn (-y^2)^n=\sumn (-1)^n y^{2n}
\] to obtain \[
\tan^{-1}x=\int_0^x \frac{1}{1+y^2}\,dy = \sumn (-1)^n\int_0^x y^{2n}\,dy= \sumn (-1)^n\frac{x^{2n+1}}{2n+1}.
\] If we replace $x$ by $x^3/6$, we obtain \[
f(x)=\tan^{-1}\frac{x^3}{6} = \sumn (-1)^n\frac{1}{2n+1} \brac{\frac{x^3}{6}}^{2n+1}
\]If we compare the coefficients of $x^9$ on both sides, we obtain \[
\frac{f^{(9)}(0)}{9!} = -\frac{1}{3}\brac{\frac{1}{6}}^3\implies f^{(9)}(0)=-\frac{9!}{3\cdot 6^3}=-560.
\] Which can be verified here.
In this example we will adopt the following product sign:\[
a_1a_2a_3\cdots a_n=\prod_{i=1}^n a_i.
\] The standard property is that for every $k\in \R$, one has \[
\prod_{i=1}^n (ka_i ) = ka_1 \cdot ka_2\cdots ka_n = k^n a_1a_2\cdots a_n =k^n \prod_{i=1}^n a_i.
\] This is the only standard fact that we need to know for problem 7.
Problem 7. Denote $f(x)=(1-4x)^{-1/2}$, then we expand $f$ by Taylor series expansion:\[
f(x)=\sum_{k=0}^\infty \frac{f^{(k)}(0)}{k!}x^k.
\] Thus our main task is to compute $f^{(k)}(0)$, which we proceed as follows:\begin{align*}
f'(x)&=\brac{-\frac{1}{2}}(1-4x)^{-1/2-1}(-4)\\
f''(x)&= \brac{-\frac{1}{2}}\brac{-\frac{1}{2}-1}(1-4x)^{-1/2-2}(-4)^2\\
&\vdots\\
f^{(k)}(x)&=\brac{-\frac{1}{2}}\brac{-\frac{1}{2}-1}\cdots \brac{-\frac{1}{2}-(k-1)}(1-4x)^{-1/2-k}(-4)^k,
\end{align*} so generally we conclude $f(0)=1$ and for $k\ge 1$,\begin{align*}
f^{(k)}(0)&=\left[\prod_{i=0}^{k-1}\brac{-\frac{1}{2}-i}\right](-4)^k\\
&=(-4)^k\prod_{i=0}^{k-1}\frac{(-1)(2i+1)}{2}\\
&=\frac{(-4)^k}{2^k} (-1)^k\prod_{i=0}^{k-1}(2i+1)\\
&=2^k \frac{\color{blue}{\prod_{i=1}^k(2i)}\prod_{i=0}^{k-1}(2i+1)}{\color{blue}{\prod_{i=1}^k(2i)}}\\
&=\frac{(2k)!}{k!}.
\end{align*} So the coefficient of our Taylor expansion of $x^k$ is \[
\frac{f^{(k)}(0)}{k!}=\frac{(2k)!}{k!k!}=\binom{2k}{k}.
\]
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