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%  The Alternating Series Test                     %
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\problemgrouptitle{The Alternating Series Test}

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

\addtoanswerkey{\begin{multicols}{2}}
  
\item $\displaystyle \sum_{n=1}^{\infty} (-1)^n \frac{n+1}{n^2+5}$
\answer{
convergent
}

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

\item $\displaystyle \sum_{n=1}^{\infty}(-1)^{n-1} \ln \left( \frac{5n^2+2}{4n^2+3} \right)$
\answer{
divergent
}

\item $\displaystyle \sum_{n=1}^{\infty}(-1)^n \frac{e^{2n}}{n^2}$
\answer{
divergent
}

\item $\displaystyle \sum_{n=1}^{\infty} (-1)^{n+1}\frac{n+5}{4^n}$
\answer{
convergent
}

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

\item $\displaystyle \sum_{n=1}^{\infty} (-1)^{n+1}  \frac{\sqrt{3n+10}}{5n+7}$
\answer{
convergent
}

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

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

\item $\displaystyle \sum_{n=1}^{\infty} \frac{\cos(n\pi)}{n}$
\answer{
convergent
}

\addtoanswerkey{\end{multicols}}

\end{problemblock}

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