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\problemgrouptitle{Sequences and Series Review}

\begin{problemblock}{Determine whether the given sequence is convergent or divergent.}

\item $\displaystyle a_n=\frac {2n}{3n+5}$
\answer{
convergent 
}

\item $\displaystyle a_n=\ln \left(\frac {3} {2n+1}\right)$
\answer{
divergent 
}

\item $\displaystyle a_n=\sqrt{4n^2+5n}-2n$
\answer{
convergent 
}

\end{problemblock}

\begin{problemblock}{Determine whether the infinite sequence is increasing, decreasing or not monotonic.}

\item $\displaystyle a_n=\frac{n+1}{5n+3}$
\answer{
decreasing
}

\item $\displaystyle a_n=n^2 e^{-6n}$
\answer{
decreasing
}

\end{problemblock}

\item Consider the series $\displaystyle\sum_{k=1}^\infty a_k=\sum_{k=1}^{\infty} \ln\left(\frac{k+1}{k}\right)$\,.
  \begin{enumerate}
  \item Find the limit $\displaystyle\lim_{k\to\infty}a_k$.  What does this say about the convergence of the series $\sum a_k$?
  \item Find the $n^{\text{th}}$ partial sum $\displaystyle S_n=\sum_{k=1}^na_k$ of the series.\\ (Hint: Expand the logarithm to get a telescoping sum.)
  \item Evaluate $\displaystyle\lim_{n\to\infty}S_n$.  Does the series $\displaystyle\sum_{k=1}^\infty a_k$ converge or diverge?
  \end{enumerate}
  \answer{
    \begin{enumerate}
    \item $\displaystyle\lim_{k\to\infty}a_k=\lim_{k\to\infty}\ln\left(\frac{k+1}{k}\right)=\ln\left(\lim_{k\to\infty}\frac{k+1}{k}\right)=\ln(1)=0$\,. Since the terms converge to zero this tells us nothing about the convergence of the series $\sum a_k$.
    \item $\displaystyle S_n=\sum_{k=1}^n\ln\left(\frac{k+1}{k}\right)=\sum_{k=1}^n\left[\ln(k+1)-\ln(k)\right]$\\[1em]
          $=[\cancel{\ln(2)}-\underbrace{\ln(1)}_{=0}]+[\cancel{\ln(3)}-\cancel{\ln(2)}]+[\cancel{\ln(4)}-\cancel{\ln(3)}]+\ldots+[\ln(n+1)-\cancel{\ln(n)}]=\ln(n+1)$
    \item $\displaystyle\lim_{n\to\infty}S_n=\lim_{n\to\infty}\ln(n+1)=\infty$.  Series diverges.
    \end{enumerate}
    }

\addtoanswerkey{\begin{multicols}{2}}
    
\begin{problemblock}[3]{Determine whether the infinite series is convergent or divergent.}

\item $\displaystyle \sum_{n=1}^{\infty} \frac{4n}{n^4+7}$
\answer{
convergent
}

\item $\displaystyle \sum_{n=1}^{\infty} \frac{2n}{n^2+10}$
\answer{
divergent
}

\item $\displaystyle \sum_{n=1}^{\infty} \frac{7+6^n}{4^n}$
\answer{
divergent
}

\item $\displaystyle \sum_{n=1}^{\infty} \left [\frac{5}{n(n+1)}-2^{-n} \right]$
\answer{
convergent
}

\item $\displaystyle \sum_{n=1}^{\infty} \frac{10}{4n+3}$
\answer{
divergent
}

\item $\displaystyle \sum_{n=1}^{\infty} n^3 e^{-2n^4}$
\answer{
convergent
}

\item $\displaystyle \sum_{n=1}^{\infty} \frac {5n^2+6}{n^2 \; e^n}$
\answer{
convergent
}

\item $\displaystyle \sum_{n=1}^{\infty} \; (-1)^n \;  \frac {\sqrt{2n+3}}{6n+7}$
\answer{
convergent
}

\item $\displaystyle \sum_{n=1}^{\infty} \;  \frac {(-3)^n}{n^3}$
\answer{
divergent
}

\item $\displaystyle \sum_{n=1}^{\infty}  \;(-1)^n\;  \frac {n+3}{(n+1)(n+2)}$
\answer{
convergent
}

\item $\displaystyle \sum_{n=1}^{\infty} \;(-1)^{n-1}\;  \frac {2^n}{n^3+4}$
\answer{
divergent
}

\item $\displaystyle \sum_{n=1}^{\infty} \; \frac {200-n}{n!}$
\answer{
convergent
}

\item $\displaystyle \sum_{n=1}^{\infty} \; (-1)^n\;  \frac {3}{n^2 \; \left( \ln n \right  )^4}$
\answer{
convergent
}

\item $\displaystyle \sum_{n=1}^{\infty} \; \frac {(n+2)!}{e^{n^2}}$
\answer{
convergent
}

\item $\displaystyle \sum_{n=1}^{\infty} \; \left( \frac {3n-5}{5n+6}\right)^n$
\answer{
convergent
}

\end{problemblock}

\addtoanswerkey{\end{multicols}}

\begin{problemblock}[2]{Find the interval and the radius of convergence of the power series.}

\item $\displaystyle \sum_{n=1}^{\infty} \; (-1)^n\;   \frac{(x-4)^{3n}}{(3n+1)!}$
\answer{
$\displaystyle  I=\left(-\infty, \infty \right),\  R=\infty$
} 

\item $\displaystyle \sum_{n=0}^{\infty} \;   \frac{1}{n^2} \; (x-7)^{n}$
\answer{
$\displaystyle  I=\left[ 6, 8 \right],\  R=1$
} 

\item $\displaystyle \sum_{n=0}^{\infty} \;   \frac{1}{\sqrt[3] n} \; (x-10)^{n}$
\answer{
$\displaystyle  I=\left[ 9, 11 \right),\  R=1$
}

\item $\displaystyle \sum_{n=0}^{\infty} \;   \frac{1}{n^2+3} \; (2x+9)^{n}$
\answer{
$\displaystyle  I=\left[ -5, -4 \right],\  R=\frac 12$
} 

\end{problemblock}

\begin{problemblock}{Find the Maclaurin series for $f(x)$ and state the radius of  convergence.}

\item $\displaystyle f(x)=\sin^2 x$
\answer{
$\displaystyle \frac 12-\frac 12 \sum_{n=0}^{\infty}(-1)^n \frac {2^{2n}}{(2n)!}x^{2n},\   R=\infty$
}

\item $\displaystyle f(x)=x e^{-5x}$
\answer{
$\displaystyle \sum_{n=0}^{\infty} \; (-1)^n \;\frac {5^n}{n!} x^{n+1},\   R=\infty$
}

\end{problemblock}

\begin{problemblock}{Find the Taylor  series for $f(x)$ centered at the given value of $a$.}

\item $\displaystyle f(x)=e^{-5x},\  a=2$
\answer{
$\displaystyle \sum_{n=0}^{\infty} \; (-1)^n \;\frac {5^n\; e^{-10}}{n!} (x-2)^{n}$
}

\item $\displaystyle f(x)=\ln (x+2),\  a=0$
\answer{
$\displaystyle \ln 2+\frac{1}{2} \; x-\frac{1}{8}  \;x^2+\frac{1}{24} \;x^3+\ldots$
}

\end{problemblock}

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