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\begin{document}

\noindent
Does $\displaystyle \sum_{n=1}^\infty \frac{n^2+1}{n^3+7}$
diverge, converge absolutely, or converge conditionally?

\subsection*{Solution 1}

$\displaystyle \sum_{n=1}^\infty \frac1n$ is a $p$-series with $p=1$. Since $p \leq 1$, the series $\displaystyle \sum_{n=1}^\infty \frac1n$ diverges by the $p$-series test. If we use $a_n = \frac{n^2+1}{n^3+7}$ and $b_n = \frac1n$, then
\begin{align*}
\lim_{n \to \infty} \frac{a_n}{b_n}
&= \lim_{n \to \infty} \frac{n^2+1}{n^3+7} \cdot \frac{n}{1}\\
&= \lim_{n \to \infty} \frac{n^3+n}{n^3+7} \\
&= \lim_{n \to \infty} \frac{3n^2+1}{3n^2} \text{ using L'hopital}\\
&= \lim_{n \to \infty} \frac{6n}{6n} \text{ using L'hopital}\\
&= \lim_{n \to \infty} 1\\
&= 1
\end{align*}
So by the Limit Comparison Test, the series $\displaystyle \sum_{n=1}^\infty \frac{n^2+1}{n^3+7}$ diverges.


\subsection*{Solution 2}

We are looking for a fixed $K > 0$ such that 
\[ K \cdot  \frac{n^2+1}{n^3+7} \geq \frac1n\]
By multiplying both sides by $n(n^3+7)$, we have
\[ K(n^3 + n) \geq n^3+7\]
so
\[ Kn^3 + Kn \geq n^3+7\]
so
if we pick $K=8$, then we'd have
\[ 8n^3 + 8n \geq n^3+7\]
is likely true because 
\[ n^3+ 7n^3 + 8n \geq n^3+7\]
with the $n^3$'s compared together, and $7n^3$ is greater than $7$. So we are ready to present starting from true inequalities.

With the scratch work above done, since $7n^3 \geq 7$ for $n \geq 1$, we have
\[ n^3+ 7n^3 + 8n \geq n^3+7\]
\[ 8n^3 + 8n \geq n^3+7\]
\[ 8(n^3 + n) \geq n^3+7\]
and dividing both sides by $8n(n^3+7)$ we get
\[\frac{n^2+1}{n^3+7} \geq \frac1{8n}\]
Since $\displaystyle \sum_{n=1}^\infty \frac1{8n} = \frac18\sum_{n=1}^\infty \frac1{n}$ diverges by the $p$-test, the series 
$\displaystyle \sum_{n=1}^\infty \frac{n^2+1}{n^3+7}$ diverges by the Direct Comparison Test.


\end{document}%%%%%%%%%%%%%%%%%

\begin{align*}
L&=\lim_{n \to \infty} \sqrt[n]{|a_n|}\\
&= \lim_{n \to \infty} \sqrt[n]{\left| \right|}\\
\end{align*}

\begin{align*}
L&=\lim_{n \to \infty} \left|\frac{a_{n+1}}{a_n}\right|\\
&= \lim_{n \to \infty} \left| \right|\\
\end{align*}

\begin{align*}
\lim_{n \to \infty} a_n
&= \lim_{n \to \infty} \\
\end{align*}


Since $\sum |a_n| = \sum a_n$, the series $\displaystyle \sum_{n=1}^\infty AAAAAAAAAAAAAA$ converges absolutely.

Since $|r| < 1$, the series ...  converges by the Geometric Series Test.

Since $|r| \geq 1$, the series ...  diverges by the Geometric Series Test.

The function $f(x)=\frac{}{}$ is continuous, positive, and decreasing on $[1,\infty)$.

\subsection*{Solution}

