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\begin{center}
\textbf{\large Math 148 Assignment 6}\\[1ex]
\textbf{Due  2:00 p.m.\ Friday, March 20 in the Math 148 dropbox.}  
\end{center}

\vskip 1em 
\begin{Ex}

\item Find the radius of convergence for the following series, and evaluate the function.
\begin{parts}
\item  $f(x) = \dsum_{n=0}^\infty (n^2+n)x^n$
\hspace{23mm} (b) $g(x) = \dsum_{n=0}^\infty \dfrac{x^{2n+1}}{(2n+1)!}$
\item[(c)] $h(x) = \dsum_{n=2}^\infty \dfrac{(-1)^n x^n}{n^2-n}$
\hspace{26mm} (d)  Evaluate $\dsum_{n=2}^\infty \dfrac{(-1)^n}{n^2-n}$. Justify!
\end{parts}


\item Suppose that $f(x) = \dsum_{j=0}^\infty a_j x^j$ and $g(x) = \dsum_{k=0}^\infty b_k x^k$ \vspace{-1ex}
have positive radii of convergence $R_1$ and $R_2$ respectively. 
Let $c_n = \dsum_{j=0}^n a_j b_{n-j}$ for $n\ge0$; and let $R = \min\{R_1, R_2\}$. \\[-2ex]
\begin{parts}
\item Define $h(x) = \dsum_{n=0}^\infty c_n x^n$.  Prove that $h(x) = f(x) g(x)$ on $(-R,R)$. 
\item Give an example where $h$ has radius of convergence strictly greater than $R$. \vspace{.5ex}
\end{parts}


\item A function defined on $\bR$ satisfies the DE $f'(x) = 2xf(x) + 4x$ and $f(0) = 1$. \\
Find a power series for the solution, and evaluate it.

\item 
\begin{parts}
\item Verify that $4 \tan^{-1}(\frac15) - \tan^{-1}(\frac1{239}) = \dfrac{\pi}4$.

\item Use (a) and Taylor polynomials to calculate $\pi$ to 6 decimals of accuracy with error estimates. \vspace{-.4ex}
\end{parts}


\item Use Taylor polynomials  and \textit{big $O$}  arguments to compute these limits (no L'H\^opital rule).\vspace{-1.5ex}
\begin{parts}
\item $\dlim_{x\to 0} \dfrac{x \cos x - \sin x}{x \sin^2 x}$
\hspace{30mm} (b) $\dlim_{x\to 0} \dfrac{e^x + e^{-x} - 2}{x^2}$  \vspace{.8ex}
\item[(c)] $\dlim_{x\to0}\ \cot^2 x - \dfrac1{x^2}$ 
\hspace{34mm} (d) $\dlim_{x\to 0} \Big( \dfrac{\tan x} x \Big)^{1/x^2}$ \hint take the log. 
\end{parts}



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\item 
\begin{parts}
\item  Let $X_n = \sum_{k=1}^n \frac1k - \ln(n+1)$ for $n\ge1$.
Express $X_n$ as an area. \vspace{.3ex}
Hence show that $X_n < X_{n+1} < 1$ for all $n\ge1$.
Deduce that this sequence has a limit, usually called $\gamma$.

\item Use (a) to obtain a useful expression for $\sum_{k=1}^{p_n} \frac1{2k-1} -  \sum_{k=1}^{q_n} \frac1{2k}$.\\
\hint $\sum_{k=1}^p \frac1{2k-1} = \sum_{k=1}^{2p} \frac1{k} - \sum_{k=1}^p \frac1{2k}$.

\item Consider a \textit{simple rearrangement} of the alternating harmonic series $\sum_{n=1}^\infty \frac{(-1)^{n+1}}n$ that contains the odd terms in order
and the even terms in order, but  the first $n$ term of the series consists of $p_n$ odd terms and $q_n$ even terms.
Prove that the series converges if and only if the limit $\dlim_{n\to\infty} \dfrac{p_n}{q_n} = \alpha$ exists.
Express the limit as a function of $\alpha$.
\end{parts}

\end{Ex}
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