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\flushleft\underline{Quiz 3 --- MATH 1586 Spring 2018} \\
\begin{enumerate}
\item Apply the ratio test to $\displaystyle\sum_{k=1}^{\infty} \dfrac{2^k}{k!}$ to conclude whether or not the series converges. \\
\textit{Solution}: In this problem, $a_k=\dfrac{2^k}{k!}$. So, compute
$$\begin{array}{ll}
\displaystyle\lim_{k \rightarrow \infty} \left| \dfrac{a_{k+1}}{a_k} \right| &= \displaystyle\lim_{k\rightarrow \infty} \left| \dfrac{\frac{2^{k+1}}{(k+1)!}}{\frac{2^k}{k!}} \right| \\
&= \displaystyle\lim_{k \rightarrow \infty} \left| \dfrac{k! 2^{k+1}}{2^k (k+1)!} \right| \\
&=\displaystyle\lim_{k \rightarrow \infty} \left| \dfrac{2}{k+1} \right| \\
&= 0.
\end{array}$$
Therefore, by the ratio test, since $0<1$, we conclude that the series converges. \\
(\textit{note: this series happens to converge to $e^2-1$})

\item Apply the ratio test to $\displaystyle\sum_{k=1}^{\infty} 2^k$ to conclude whether or not the series converges. \\
\textit{Solution}: In this problem, $a_k = 2^k$. So, compute
$$\begin{array}{ll}
\displaystyle\lim_{k \rightarrow \infty} \left| \dfrac{a_{k+1}}{a_k} \right| &= \displaystyle\lim_{k \rightarrow \infty} \left| \dfrac{2^{k+1}}{2^k} \right| \\
&=\displaystyle\lim_{k \rightarrow \infty} |2| \\
&= 2.
\end{array}$$
Since $2>1$, we conclude by the ratio test that the series diverges.
\end{enumerate}
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