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\title{Math 166: Honors Calculus II\\
Exam II Solutions}
\author{April 8, 1999}

\begin{document}

\maketitle

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

\nextproblem

3. a) Integrating the formula $\frac{1}{1+x} = 1 - x + x^2 + o(x^3)$,
gives $\log(1+x) = x - \frac{1}{2}x^2 + \frac{1}{3}x^3 + o(x^{3})$ so
$x^3\log(1+x) = x^4 - \frac{1}{2}x^5 + \frac{1}{3}x^6 + o(x^{6})$.

b) Using the rule $\frac{1}{1-u} = 1+u+o(u)$ with $u=x^2-x^3+o(x^3)$
we get
  \begin{eqnarray*}
  \frac{1+x+x^2+o(x^3)}{1-x^2+x^3+o(x^3)} &=&
  (1+x+x^2+o(x^3))(1+x^2-x^3+o(x^3) + o(x^2-x^3+o(x^3))\\
  &=& (1+x+x^2+o(x^3))(1+x^2+o(x^2)) = 1+x+2x^2+o(x^2)
  \end{eqnarray*}

c) The Lagrange form of the remainder is $E_nf(x) =
\frac{1}{(n+1)!}f^{(n+1)}(c)x^{n+1}$ for some $c$ in $[-1,1]$ (between 0
and $x$). Since $f(x) = \sin(x)$, the derivatives of $f(x)$ are either
$\pm\sin(x)$ or $\pm\cos(x)$, so $|f^{(n+1)}(x)| \le 1$. Thus,
$|E_nf(x)| \le \frac{1}{(n+1)!}|x|^{n+1} \le \frac{1}{(n+1)!}$ since,
by assumption, $|x|<1$.  Therefore $|E_n| < 10^{-3}$ if $(n+1)! > 1000$
which occurs for $n\ge 6$ ($6! = 720$ and $7! = 5040$).

d) By a familiar algebraic identity, the partial sums are $s_n =
\sum_{k=0}^n r^k = \frac{1}{1-r}-\frac{r^{n+1}}{1-r}$. Since $|r|^{n+1}
\to 0$ as $n\to \infty$ when $|r|<1$, the limit of the partial sums is
$\lim_{n\to\infty}s_n = \frac{1}{1-r}$.

\nextproblem

4.  a) $\lim_{x\to 0}(1+h(x))^{1/h(x)}
= \exp[\lim_{x\to0}\log(1+h(x))/h(x)] =
\exp[\lim_{x\to0}\frac{h'(x)/(1+h(x))}{h'(x)}] =
\exp[\lim_{x\to0}\frac{1}{1+h(x)}] = e$.  We could also solve this
problem assuming only that $h(x)$ is continuous by substituting $u =
h(x)$ and $u\to0$ for $x\to0$: $\exp[\lim_{x\to0}\log(1+h(x))/h(x)]
=\exp[\lim_{u\to0}\log(1+u)/u] =\exp[\lim_{u\to0}\frac{1}{1+u}] = e$ (use
L'H\^opital's rule, taking derivatives with respect to the variable $u$!).


b) Use L'Hopital's rule twice: $\lim_{x\to 0} \frac{e^{x^2}-\cos(x)}{x^2}
=\lim_{x\to 0} \frac{2xe^{x^2}+\sin(x)}{2x} =\lim_{x\to 0}
\frac{(2+4x^2)e^{x^2}+\cos(x)}{2} = \frac{3}{2}$.  We could also use
$o$-notation: $\lim_{x\to 0} \frac{e^{x^2}-\cos(x)}{x^2} =\lim_{x\to 0}
\frac{(1+x^2+o(x^3)) -(1-\frac{1}{2}x^2+o(x^3)}{x^2} =\lim_{x\to 0}
\frac{3}{2}+o(x) = \frac{3}{2}$.


c) $\lim_{x\to\infty} \sqrt{2x+x^2} - \sqrt{x+x^2} = \lim_{x\to\infty}
\frac{(2x+x^2)-(x+x^2)}{\sqrt{2x+x^2}+\sqrt{x+x^2}} = \lim_{x\to\infty}
\frac{1}{\sqrt{\frac{2}{x}+1}+\sqrt{\frac{1}{x}+1}} =\frac{1}{2}$.  We
could also substitute $x = \frac{1}{y}$: $\lim_{x\to\infty} \sqrt{2x+x^2}
- \sqrt{x+x^2} =\lim_{y\to0^+} \frac{\sqrt{2y+1} - \sqrt{y+1}}{y}
=\lim_{y\to0^+} \frac{(2y+1)-(y+1)}{y(\sqrt{2y+1}+\sqrt{y+1})}
=\lim_{y\to0^+} \frac{1}{\sqrt{2y+1}+\sqrt{y+1}} = \frac{1}{2}$.


d) $\lim_{n\to\infty}\frac{(n + (-1)^n)(n+1)}{3n^2} =
\lim_{n\to\infty}\frac{n^2 + (1+(-1)^n)n +1}{3n^2} =
\lim_{n\to\infty}\left(\frac{1}{3} + \frac{1+(-1)^n}{3n} +\frac{1}{3n^2}\right) =
\frac{1}{3}$

\nextproblem

5.
a) $\sum_{n=0}^\infty \frac{a^{n} + (-1)^n}{b^{2n}} = \sum_{n=0}^\infty
\Big(\frac{a}{b^2}\Big)^n + \Big(\frac{-1}{b^2}\Big)^n = \frac{1}{1-a/b^2}
+ \frac{1}{1+1/b^2}$ (linearity and geometric series).

b) $\sum_{n=1}^\infty \frac{2n + 1}{n^2(n+1)^2} =
\sum_{n=1}^\infty \frac{1}{n^2}-\frac{1}{(n+1)^2} = \frac{1}{1^2} -
\lim_{n\to\infty}\frac{1}{(n+1)^2} = 1$ (telescoping sum).

c) $\sum_{n=0}^\infty \frac{n}{n+1}x^n
=\sum_{n=0}^\infty \Big(1-\frac{1}{n+1}\Big)x^n
=\sum_{n=0}^\infty x^n-\sum_{n=0}^\infty\frac{1}{n+1}x^n
=\frac{1}{1-x} - \frac{1}{x}\sum_{n=0}^\infty\frac{1}{n+1}x^{n+1}\newline
=\frac{1}{1-x} - \frac{1}{x}\int_0^x\sum_{n=0}^\infty t^n\,dt
=\frac{1}{1-x} - \frac{1}{x}\int_0^x\frac{1}{1-t}\,dt
=\frac{1}{1-x} + \frac{1}{x}\log(1-x)$.
We could also do the following:
$\sum_{n=0}^\infty \frac{n}{n+1}x^n
=x\frac{d}{dx}\sum_{n=0}^\infty \frac{1}{n+1}x^n
=x\frac{d}{dx}\Big[\frac{1}{x}\int_0^x \sum_{n=0}^\infty t^n dt \Big]
=x\frac{d}{dx}\Big[\frac{1}{x}\int_0^x \frac{1}{1-t} dt \Big]
=x\frac{d}{dx}\Big[-\frac{1}{x}\log(1-x)\Big]
=x[\frac{1}{x^2}\log(1-x)+\frac{1}{x(1-x)}]
=\frac{1}{x}\log(1-x)+\frac{1}{1-x}$

\nextproblem

6.
a) Since $\frac{1}{n!}<\frac{1}{2^{n-1}}$ for integers $n\ge1$, and
$\sum_{n=1}^{\infty}\frac{1}{2^{n-1}}$ converges (geometric series),
the Comparison Test implies that $\sum_{n=0}^\infty\frac{1}{n!}$
converges.  We could also compare $\frac{1}{n!} \le \frac{1}{n(n+1)} =
\frac{1}{n} - \frac{1}{n+1}$ and $\sum_{n=1}^{\infty} \Big(\frac{1}{n}
- \frac{1}{n+1}\Big)$ converges as a telescoping sum.

b) First we note that $\frac{\sqrt{n+1}}{n^2}$ is asymptotically equal to
$\frac{1}{n^{3/2}}$: $\lim_{n\to\infty}\frac{\sqrt{n+1}/n^2}{1/n^{3/2}}
=\lim_{n\to\infty}\sqrt{1+\frac{1}{n}} = 1$. We could also
compare $\frac{\sqrt{n+1}}{n^2} \le \frac{\sqrt{2n}}{n^2}
= \frac{\sqrt{2}}{n^{3/2}}$.  Since $\sum_{n=1}^{\infty}
\frac{1}{n^{3/2}}$ converges (by the Integral Test, for example), the
series $\sum_{n=1}^\infty\frac{\sqrt{n+1}}{n^2}$ converges.

c) Using the substitution $u=\log(x)$, $du = \frac{1}{x}\,dx$,
we find
  \[ \int_2^n \frac{1}{x\log(x)}\,dx = \int_{\log(2)}^{\log(n)}
  \frac{1}{u}\,du = \log|\log(n)|-\log|\log(2)|\to\infty\ \mbox{as}\
  n\to\infty \]
Therefore the series $\sum_{n=2}^\infty\frac{1}{n\log(n)}$ diverges by
the Integral Test.

\nextproblem

\end{document}
