Problem 5:
Given that
\begin{equation} \label{eq:poly}
\sum_{n=1}^{\infty}{a_n}
\end{equation}
converges state which of the following converges, may converge or does not converge.
\begin{equation} \label{eq:poly}
\sum_{n=1}^{\infty}\frac{a_n}{n^2}
\end{equation}
We can use the limit comparison test. Let
\[ b_n=\frac{a_n}{n^2} \]
Then,
\[ \frac{b_n}{a_n} = \frac{1}{n^2} \]
Since
\begin{equation} \label{eq:poly}
\lim n\rightarrow \infty \frac{1}{n^2} =0
\end{equation}
We can conclude that
\begin{equation} \label{eq:poly}
\sum_{n=1}^{\infty}\frac{a_n}{n^2}
\end{equation}converges
We can also use the comparison test by noting that
\begin{equation} \label{eq:poly}
\sum_{n=1}^{\infty}\frac{a_n}{n^2} < \sum_{n=1}^{\infty}{a_n}
\end{equation}
to conclude that
\begin{equation} \label{eq:poly}
\sum_{n=1}^{\infty}\frac{a_n}{n^2}
\end{equation}converges
Problem 7
To determine the radius of convergence we can apply the ratio test to
\begin{equation} \label{eq:poly}
\sum_{n=1}^{\infty}{(n+2)}{x^n}
\end{equation}
$\frac{a_{n+1}}{a_n} = \frac{(n+3)x^{n+1}}{(n+2)x^n} =\frac{(n+3)}{(n+2)}x $
Since $\lim n\rightarrow \infty \frac{n+2}{n+1} =1$ we need that $|x|<1$ for convergence.
To get the interval of convergence we need to check the two endpoints of $-1,1$
If $x = -1$, the series will alternate and increase since $n+2$ is increasing. The same is true for $x=1$ the terms are increasing in size.
Conclusion: Interval of convergence is $(-1,1)$
Problem 10
Apply the root test to: $\sum_{n=1}^{\infty}\frac{1}{5^n}$
$\sqrt[n]{a_n}=\;\sqrt[n]{\frac{1}{5^n}}\;=\frac{1}{5}\;<1$ so converges.
Problem 12
$ln|1-x|\;=\int \frac{1}{1-x}dx$ so to get the power series for $\ln|1-x|$ we can integrate the series for $\frac{1}{1-x}$ which is
\begin{equation} \label{eq:poly}
\frac{1}{1-x}\;=\sum_{n=0}^{\infty}x^n
\end{equation}
