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\begin{document}
\section*{Quiz 7}
\section*{Name:\hspace{4cm}\;}
\subsection*{Problem 1} Find the center,  radius and interval of convergence for this power series.
\[\sum_{n=0}^\infty n!(x-5)^n\]
\solution We use the ratio test.
\begin{align*}
 \lim_{n\to\infty} \frac{(n+1)!(x-5)^{n+1}}{(n)!(x-5)^n}=&\lim_{n\to\infty} (n+1) (x-5)
\end{align*}
This limit is finite only when $x=5$. Therefore we see that this series converges only when $x=5.$ So the series is centered at $5$,  has a radius of convergence of $0$,  and an interval of convergence $[5, 5]$.\\\vspace{2cm}
\subsection*{Problem 2} Find a power series that fits this function. Then find  radius and center of convergence of this power series. 
\[\frac{(x+2)^3}{1+4x+x^2}\]
\solution  Here is how I would solve the problem. Follow the arrows!
\[\begin{tikzcd}
   \frac{(x+2)^3}{1+4x+x^2}\arrow{d}[left]{\text{We pull out a term and start to complete the square}}&=&\sum_{n=0}^\infty  -3^{-(n+1)}(x+2)^{2n+3}\\
   (x+2)^3\frac{1}{1-4+4+4x+x^2}\arrow{d}[left]{\text{Completing the square}}&=&	-\frac{(x+2)^3}{3}\sum_{n=0}^\infty \left(\frac{1}{3^n}\right)^n (x+2)^{2n}\arrow{u}[right]{\text{Bringing in the terms from the outside}}\\
  (x+2)^3\frac{1}{-3+(x+2)^2}\arrow{d}[left]{\text{Substituting $3u=(x+2)^2$}}&=&	-\frac{(x+2)^3}{3}\sum_{n=0}^\infty (\frac{1}{3}(x+2)^2)^n\arrow{u}[right]{\text{Breaking up the power}}\\
  - \frac{(x+2)^3}{3}\left(\frac{1}{1-u}\right)\arrow{rr}{=}&&					-\frac{(x+2)^3}{3}\sum_{n=0}^\infty u^n\arrow{u}[right]{\text{Undoing the substitution for $u$}}
  \end{tikzcd}
\]


And now we just need to convert the sum $\sum_{n=0}^\infty  -3^{-(n+1)}(x+2)^{2n+3}$ into a proper power series. Letting
\[c_m=\left\{\begin{array}{ll}
              - (3)^{-\frac{m-1}{2}}& \text{$m$ is odd and greater than 2}\\
              0 &\text{Otherwise}
             \end{array}\right.\]
We have that $\sum_{m=0}^\infty c_m x^m $ is the desired power series. It is centered at $x=-2$,  and has a radius of convergence of $\sqrt 3$
\newpage

\subsection*{Problem 3} Evaluate the following sum.
\[\frac{1}{4\cdot 1}+\frac{1}{16\cdot 2}+\frac{1}{64\cdot 3}+\frac{1}{256\cdot 4}+\ldots+\frac{1}{2^{2n}\cdot n}+\ldots\]
\solution We notice that this sum can be written 
\[\begin{tikzcd}
  \sum_{n=1}^\infty \frac{1}{2^{2n}\cdot n}\arrow{d}[left]{\text{Rearranging the terms}}		&=&-\ln(3/4)\\
    \sum_{n=1}^\infty \frac{(1/4)^n}{n}\arrow{d}[left]{\text{Letting $x=1/4$}}&=&-\ln(1-1/4)\arrow{u}[right]{\text{Simplifying}}\\
   \sum_{n=1}^\infty \frac{(x)^n}{n}\arrow{d}[left]{\text{Differentiating the series}}&=&-\ln(1-x)+C\arrow{u}[right]{\text{Solving for $C$ shows that $C=0$. Putting the $x$ back in.}}\\
   \sum_{n=1}^\infty x^{n-1}\arrow{rr}{=}		&&\frac{1}{1-x}\arrow{u}[right]{\text{Now integrating to undo the differentiation}}
  \end{tikzcd}
\]


\end{document}
