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\title{Math 166: Honors Calculus II\\
Exam II Solutions}

\date{April 20, 2000}

\begin{document}

\maketitle

\bigskip

Problems 1 and 2 are definitions and statements of theorems that
can be found in the text.

\nextproblem

3. a)
\begin{eqnarray*}
\frac{x^2\sin(x) - x\sin(x^2)}{\sin(x^5)}&=&
\frac{x^2\big(x-(1/3!)x^3+o(x^4)\big)-x\big(x^2-(1/3!)(x^2)^3+o((x^2)^4)\big)}
    {x^5+o((x^5)^2)}\cr
&=&\frac{x^3-(1/6)x^5+o(x^6)-x^3+(1/6)x^7 + o(x^9)}
              {x^5+o(x^{10})}\cr
&=&\frac{-(1/6)x^5+o(x^6)}{x^5 + o(x^{10})}
=\frac{- 1/6 + o(x)}{1 + o(x^5)} \longrightarrow -\frac{1}{6}
\quad\mbox{as $x\to0$}
\end{eqnarray*}

b) Let $x = 1/n$, then
$(n+1)^c-n^c= (1/x+1)^c - (1/x)^c = \frac{(1+x)^c - 1}{x^c}$. So
\[\begin{array}{rcll}
\lim_{n\to\infty}(n+1)^c-n^c&=&\lim_{x\to0^+}\frac{(1+x)^c-1}{x^c}
=\lim_{x\to0^+}\frac{c(1+x)^{c-1}}{cx^{c-1}}&\textsc{[L'Hopital]}\cr
&=&\lim_{x\to0^+}\Big(\frac{x}{1+x}\Big)^{1-c} = 0&[1-c>0]\cr
\end{array}\]

c) First observe that $\lim_{x\to1} \frac{\log(x)}{1-x} = \lim_{x\to1}
\frac{1/x}{-1} = -1$ by \textsc{L'Hopital's Rule}. Therefore,
$\lim_{x\to1} x^{1/(1-x)} =\lim_{x\to1} e^{\log(x)/(1-x)} = e^{-1} = 1/e$.

\nextproblem

4. If $|x|<1$ then $\sum_{n=k}^\infty x^n = \frac{x^k}{1-x}$.

\begin{proof} Using the algebraic identity $\sum_{n=0}^p x^n =
\frac{1-x^{p+1}}{1-x}$ we may write the partial sums as
\[ s_m = \sum_{n=k}^m x^n = x^k\sum_{n=k}^m x^{n-k}
 = x^k\sum_{n=0}^{m-k} x^n = x^k\frac{1-x^{m-k+1}}{1-x}
 = \frac{x^k}{1-x} - \frac{x^{m+1}}{1-x} \]
Since $|x|<1$, the second term $|x^{m+1}/(1-x)| = |x|^{m+1}/|1-x| \to 0$ as
$m\to \infty$.  Therefore,
\[ \sum_{n=k}^\infty x^n = \lim_{m\to\infty}s_m = \frac{x^k}{1-x} -
\lim_{m\to\infty}\frac{x^{m+1}}{1-x} = \frac{x^k}{1-x} \] \end{proof}

\nextproblem

5.

a)
$\sum_{n=1}^\infty \frac{(2^n-1)(2^n+1)}{5^n}
=\sum_{n=1}^\infty \frac{4^n-1}{5^n}
=\sum_{n=1}^\infty \Big(\frac{4}{5}\Big)^n -
 \sum_{n=1}^\infty\Big(\frac{1}{5}\Big)^n
=\frac{4/5}{1-4/5} - \frac{1/5}{1-1/5} = 4 - \frac{1}{4} = \frac{15}{4}$

(geometric series)


b)
$\sum_{n=1}^\infty\frac{1}{4n^2-1}
=\sum_{n=1}^\infty\frac{1}{(2n-1)(2n+1)}
=\frac{1}{2}\sum_{n=1}^\infty\Big(\frac{1}{2n-1}-\frac{1}{2n+1}\Big)
=\frac{1}{2}\sum_{n=1}^\infty\Big(\frac{1}{2n-1}-\frac{1}{2(n+1)-1}\Big)$

This is a telescoping series that sums to
$\frac{1}{2}\Big(\frac{1}{2(1)-1} - \lim_{n\to\infty}\frac{1}{2n+1}\Big)
= \frac{1}{2}$

\newpage
\nextproblem

6.
a) Let $f(x) = \frac{1}{x^{1+1/x}} = e^{-(1+1/x)\log(x)}$. Since
\[f'(x)=\bigg(\frac{\log(x)}{x^2}-(1+1/x)\frac{1}{x}\bigg)e^{-(1+1/x)\log(x)}
= (\log(x) - x - 1)\frac{1}{x^{3+1/x}} < 0\] $f(x)$ is decreasing.
Moreover, the limit of $f(x)$ as $x\to\infty$ is clearly 0:
$0\le\lim_{x\to\infty}\frac{1}{x^{1+1/x}}\le\lim_{x\to\infty}\frac{1}{x}=0$
Since the series is alternating and the terms $\frac{1}{n^{1+1/n}}$
decrease to 0, \textsc{Leibniz's Rule} implies that the series
$\sum_{n=1}^\infty \frac{(-1)^{n-1}}{n^{1+1/n}}$ converges.
However, the absolute value series diverges by the \textsc{Limit
Comparison test} because it is comparable to the harmonic
series $\sum_{n=1}^\infty\frac{1}{n}$: $\lim_{n\to\infty}
\frac{1/n^{1+1/n}}{1/n} =\lim_{n\to\infty} \frac{1}{n^{1/n}} =
\lim_{n\to\infty}\frac{1}{e^{\log(n)/n}} = \frac{1}{e^0} = 1$. Therefore,
the convergence is conditional.

\bigskip
b) The series $\sum_{n=1}^\infty \frac{n!}{n^n}$ converges
(absolutely) by the \textsc{Ratio Test}: \[ \lim_{n\to\infty}
\frac{(n+1)!/(n+1)^{n+1}}{n!/n^n} =\lim_{n\to\infty} \frac{(n+1)
\cdot n!}{n!} \cdot\frac{n^n}{(n+1)^{n+1}} =\lim_{n\to\infty}
\Big(\frac{n}{n+1}\Big)^n =\lim_{n\to\infty} \frac{1}{(1+1/n)^n} =
\frac{1}{e} < 1\]

\bigskip
c) The series $\sum_{n=1}^\infty n(-\pi)^{-n}$ converges absolutely by
the \textsc{Root Test}: $\lim_{n\to\infty} \Big(\frac{n}{\pi^n}\Big)^{1/n}
=\lim_{n\to\infty} \frac{n^{1/n}}{\pi} = \frac{1}{\pi} < 1$.

\bigskip
d) The series $\sum_{n=2}^\infty \frac{(-1)^n}{n(\log(n))^2}$
converges absolutely by the \textsc{Integral Test}:
$\int_2^\infty\!\!\frac{1}{x(\log(x))^2}dx =
-\frac{1}{\log(x)}\Big|_2^\infty = \frac{1}{\log(2)} < \infty$.

\bigskip
e) The harmonic series $\sum_{n=2}^\infty\frac{1}{n}$ diverges
and the series $\sum_{n=2}^\infty\frac{(-1)^n}{\sqrt{n}}$
converges by \textsc{Leibniz's Rule}. Therefore, the sum of these
two series, $\sum_{n=2}^\infty\frac{\sqrt{n}+(-1)^n n}{n\sqrt{n}}
=\sum_{n=2}^\infty\Big(\frac{1}{n}+\frac{(-1)^n}{\sqrt{n}}\Big)$,
must diverge as a consequence of the linearity of series.

\nextproblem

\end{document}
