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\def\frac#1#2{{{#1}\over{#2}}}
\def\equiz{\par\noindent\vrule width\hsize height 0pt depth .1 pt
\par\global\cc=1}
\def\leaderfill{\leaders\hbox to .5em{\hss.\hss}\hfil}

\def\dquiz{\par\noindent\hbox to \hsize{\leaderfill}
\par\medskip\global\cc=1}

\long\def\stp\centerline#1 \centerline#2 \vskip10pt \noindent{%
\par\bigskip\noindent{#1} \qquad{ #2}\par\medskip\noindent}
\long\def\Stp#1\equiz{}
%%%%%%%%%%%%%%%
%################
\def\frac#1#2{{{#1}\over{#2}}}
\def\smfrac#1#2{{{\scriptstyle#1}\over{\scriptstyle#2}}}
\def\sL#1{\displaystyle\lim_{#1\to\infty}}
\def\isum#1{{\displaystyle{\sum_{n={#1}}^{\infty}}}\>}

\newcount\cc\cc=1
\def\np{%
\par\noindent{\bf\number\cc.}\ %
\global\advance\cc by1%
}



%################
\stp
%: #1
\centerline{\bf Math.~126 Quiz \#1}
\centerline{January  20, 2004}
\vskip10pt
\noindent
\def\frac#1#2{{{#1}\over{#2}}}
\itemitem{1.}
The function $$f(x) = x^5 + 5x^2 - 15x + 4$$ is increasing on the interval
$[ 1 , 3 ]$ and hence one{--}to{--}one.
Let $g(x)$ be the inverse function for $f(x)$, $x$ in $[ 1, 3 ]$.
The equation $f(x)=26$ has a solution, $x = 2$.
Use this information to find $g^\prime(26)$.
\medskip
\itemitem{2.} Write an equation for the tangent line to the graph of
$y=g(x)$ at the point $x=26$.

\dquiz
Since $f(2)=26$, $g(26)=2$.
Since $g^\prime\bigl(f(x)\bigr) = \frac{1}{f^\prime(x)}$,
$g^\prime(26) = g^\prime\bigl(f(2)\bigr) = \frac{1}{f^\prime(2)}$.
Since $f^\prime(x) = 5x^4+10x-15$, $f^\prime(2) = 5\cdot 2^4 + 10\cdot 2 - 15 =
5\cdot 16 +20 -15 = 80 + 20 -15 = 85$ and
$g^\prime(26) = \frac{1}{85}$.
\medskip
Since $g(26)=2$ and $g^\prime(26) = \frac{1}{85}$, an equation of the tangent
line is
$$y-2 = \frac{1}{85}(x-26)\ .$$
\equiz
%######################
\stp
%:#2
\centerline{\bf Math.~126 Quiz \#2}
\centerline{January  27, 2003}
\vskip10pt
\noindent
\def\frac#1#2{{{#1}\over{#2}}}
\noindent
\itemitem{1.} Solve the equation $9^{3x+4}=2$ for $x$.
An answer such as $\ln(6)-\sqrt{\ln 8}$ or whatever is fine.
\medskip
\itemitem{2.}Find the derivative with respect to $x$ of
$(\tan x)^{3x}$.
\dquiz
Since $9^{3x+4}=2$, $\ln\bigl(9^{3x+4}\bigr) = \ln 2$, so
$(3x+4) \ln 9 = \ln 2$ and so $3x+4 = \frac{\ln 2}{\ln 9}$.
Hence $3x = \frac{\ln 2}{\ln 9} - 4$ and
 $x = \frac{\frac{\ln 2}{\ln 9} - 4}{3}$.
 If you prefer, you can write this as
 $x = \frac{\ln 2 - 4\ln 9}{3\ln 9}$.
 \medskip
 Write $(\tan x)^{3x} = e^{3x \ln(\tan x)} = e^u$.
 Since $\frac{d\> e^u}{dx} = u^\prime e^u$, it behooves us to find $u^\prime$.
 Well, $\frac{d\>\bigl(3x(\ln \tan x)\bigr)}{dx} = 3(\ln \tan x) + 3x\Bigl(\frac{\sec^2 x}{\tan x}\Bigr)$, so
 $$\frac{d\>(\tan x)^{3x}}{dx} = \Bigl(
 3(\ln \tan x) + 3x\bigl(\frac{\sec^2 x}{\tan x}\bigr)\Bigr)(\tan x)^{3x}\ .
 $$
\equiz
%######################
\stp
%:3
\centerline{\bf Math.~126 Quiz \#3}
\centerline{February  10, 2004}
\vskip10pt
\noindent
\def\frac#1#2{{{#1}\over{#2}}}
\noindent
\noindent Evaluate the integral
$$\int \sec^2 \bigl(\sqrt x\>\bigr)\> dx \ .$$

\noindent{\bf Hint:} First do a substitution and
then an integration by parts.
\dquiz
The square root inside the $\sec$ does not look friendly, so let $u = \sqrt x$.
Then $du = \frac{dx}{2\sqrt{x}}$.
There is no ``extra'' square root, so let $dx = 2u du$  and
$$\int \sec^2 \bigl(\sqrt x\>\bigr)\> dx = 2\int\>u \sec^2 u\> du \ .$$
This new integral looks good for parts. 
Taking $dv$ to be the biggest chunk you can integrate suggests $dv=\sec^2 u\> du$ or
$dv = u du$.
This second try looks not so good since $v=\frac{u^2}{2}$ and the power of $u$ in the next
integral will be bigger while the derivative of the $\sec$ squared will still be trig functions.
So let's go with $dv = \sec^2 u\> du$ and $w = u$: $v = \tan u$ and $dw = du$ so
$$\int\>u \sec^2 u\> du = u\tan u - \int \tan u\>du = u \tan u - \ln\vert \sec u \vert +C\ .
$$
Hence
$$\int \sec^2 \bigl(\sqrt x\>\bigr)\> dx = 
2\Bigl(\sqrt{x} \tan\bigl(\sqrt{x}\bigr) - \ln\Big\vert \sec\bigl(\sqrt{x}\bigr)\Big\vert\Bigr)+C
$$
\equiz
%######################
\stp
%:4
\centerline{\bf Math.~126 Quiz \#4}
\centerline{February  17, 2004}
\vskip10pt
\noindent
\def\frac#1#2{{{#1}\over{#2}}}
\noindent
\itemitem{1.}Set up the initial form for the partial fraction for
$$\frac{x^3+7x^2-10x+1}{(x+2)^2(x-1)(x^2-1)^2(x^2+2x+3)^2}\ . $$
\itemitem{}{\bf Hint:}The initial form for $\frac{1}{(x+1)(x-1)}$ is $\frac{A}{x-1} + \frac{B}{x+1}$.
\medskip
\itemitem{2.}
Integrate $\displaystyle\int \frac{2x^2+2x+1}{(x+1)(x^2+1)}\ dx$.
\dquiz
The degree of the numerator is $3$ and the degree of the denominator is $11$ so there
is no need to do a polynomial long division. 
The second quadratic is irreducible since $2^2-4\cdot1\cdot 3<0$, but
$x^2-1 = (x-1)(x+1)$.
Hence
$$\eqalign{
\frac{x^3+7x^2-10x+1}{(x+2)^2(x-1)(x^2-1)^2(x^2+2x+3)^2}=
\frac{x^3+7x^2-10x+1}{(x+2)^2(x-1)^3(x+1)^2(x^2+2x+3)^2}=\cr
\frac{A}{x+2}+\frac{B}{(x+2)^2}+
\frac{C}{x-1}+\frac{D}{(x-1)^2}+\frac{E}{(x-1)^3}+
\frac{F}{x+1}+\frac{G}{(x+1)^2}+\cr
\frac{Hx+I}{x^2+2x+3}+\frac{Jx+K}{(x^2+2x+3)^2}
\ . 
\cr}$$
The order of the factors is not important, nor are the names of the variables:
you could do the $x-1$ powers first, then the quadratic, then the $x+2$ powers, etc.
\medskip
To do the integral, requires expanding 
$\frac{2x^2+2x+1}{(x+1)(x^2+1)}$ by partial fractions.
The degree of the numerator is $2$, that if the denominator $3$, so no long division is needed,
the quadratic is irreducible so
$$\frac{2x^2+2x+1}{(x+1)(x^2+1)}=\frac{A}{x+1}+\frac{Bx+C}{x^2+1}\ ,\quad{\rm so}$$
$$2x^2+2x+1=A\cdot (x^2+1)+ (Bx+C)(x+1)\ .$$
Plug in $x=-1$ and see $2(-1)^2+2(-1)+1 = A((-1)^1+1) + (Bx+C)\cdot 0$,
or $2A=1$ or $A=\frac{1}{2}$.

\noindent Hence $2x^2+2x+1=\frac{1}{2}\cdot (x^2+1)+ (Bx+C)(x+1)$, so
$4x^2+4x+2 = x^2+1 +2(Bx+C)(x+1)$ so

\noindent $3x^2+4x+1 = 2(Bx+C)(x+1) = 2Bx^2+2(B+C)x+2C$.
Hence $B=\frac{3}{2}$, $C=\frac{1}{2}$ and
$$\frac{2x^2+2x+1}{(x+1)(x^2+1)} = \frac{1}{2}\Bigl(
\frac{1}{x+1} + \frac{3x+1}{x^2+1}\Bigr)\ .$$
To do the integral, compute
$\displaystyle\int \frac{dx}{x+1} = \ln\vert x+1\vert +C$ and
$\displaystyle\int \frac{(3x+1)dx}{x^2+1}  = 3\int \frac{x\>dx}{x^2+1} +
\int \frac{dx}{x^2+1}$.
By substitution, $\displaystyle\int \frac{x\>dx}{x^2+1} = \frac{\ln(x^2+1)}{2}+C$, so
$$\int \frac{2x^2+2x+1}{(x+1)(x^2+1)}\ dx =
\frac{\ln\vert x+1\vert}{2} +  \frac{3\ln(x^2+1)}{4} + \frac{\arctan{x}}{2}+C\ .$$
\equiz
%######################
\stp
%:5
\centerline{\bf Math.~126 Quiz \#5}
\centerline{February  24, 2004}
\vskip10pt
\noindent
\def\frac#1#2{{{#1}\over{#2}}}
\noindent
Which improper integrals below converge and which diverge?
Indicate
your reasoning and be careful.
\itemitem{a)}$\displaystyle\int_{1}^\infty
\frac{1}{\root 9\of{x^{11}}}\ dx$
\itemitem{b)}$\displaystyle\int_{-\infty}^{1}
\frac{1}{\root 9\of{x^{11}}}\ dx$
\itemitem{c)}$\displaystyle\int_1^\infty \frac{\sin^2x}{\root 9\of{x^{11}}}\ dx$
\dquiz
\item{(a)} Since $x^{-\frac{11}{9}}$ is continuous on $[1,\infty)$,
$\displaystyle\int_{1}^\infty
\frac{1}{\root 9\of{x^{11}}}\ dx = \lim_{t\to\infty}\int_{1}^{t}
\frac{1}{\root 9\of{x^{11}}}\ dx = 
\lim_{t\to\infty} \frac{x^{-\frac{2}{9}}}{-\frac{2}{9}}\Bigg\vert^{t}_{1} = 
\lim_{t\to\infty} -\frac{9 t^{-\frac{2}{9}}}{2} - \biggl(-\frac{9}{2}\biggr)=
\frac{9}{2}$ since the limit as $t\to\infty$ of $t$ to a negative power, $-\frac{2}{9}$,
goes to $0$.
In particular, the integral converges.

\medskip\item{(b)}
This one is a bit tricky since $x^{-\frac{11}{9}}$ is not continuous at $0 \in (-\infty, 1]$.
Hence
$$\int_{-\infty}^{1}
\frac{1}{\root 9\of{x^{11}}}\ dx = \int_{-\infty}^{-1}
\frac{1}{\root 9\of{x^{11}}}\ dx + 
\int_{-1}^{0}
\frac{1}{\root 9\of{x^{11}}}\ dx + 
\int_{0}^{1}
\frac{1}{\root 9\of{x^{11}}}\ dx\ .$$
The $-1$ is arbitrary. You may use any negative number you like.
We suspect that the analysis at $-\infty$ will look a lot like the analysis in part (a), so
let's start with one of the other two integrals.

$$\int_{-1}^{0}
\frac{1}{\root 9\of{x^{11}}}\ dx = \lim_{t\to 0^-} \int_{-1}^{t}
\frac{1}{\root 9\of{x^{11}}}\ dx = 
\lim_{t\to0^-} -\frac{9x^{-\frac{2}{9}}}{2}\Bigg\vert^{t}_{-1} =
\frac{9}{2}\lim_{t\to0^-} t^{-\frac{2}{9}} - \biggl(-\frac{9}{2}\biggl)\ .
$$
\item{}Since $-\frac{2}{9}< 0$, $\displaystyle\lim_{t\to0^-} t^{-\frac{2}{9}} = \infty$ and this integral
diverges.
Hence the original integral also diverges.

\medskip\item{(c)}You have no hope of actually doing the integral so you can not work directly
with the definition. Instead, use a theorem.
Since $-1\leq \sin x \leq 1$, $0\leq\sin^2 x\leq 1$ and hence
$0\leq \frac{\sin^2x}{\root 9\of{x^{11}}} \leq \frac{1}{\root 9\of{x^{11}}}$.
Since $\displaystyle\int_1^\infty \frac{dx}{\root 9\of{x^{11}}}$ converges by part (a),
our theorem says 
$\displaystyle\int_1^\infty \frac{\sin^2 x\>dx}{\root 9\of{x^{11}}}$ also converges.
\equiz
%######################
%\vfill\eject\stp
%:6
\centerline{\bf Math.~126 Quiz \#6}
\centerline{March  2, 2004}
\vskip10pt
\noindent
\def\frac#1#2{{{#1}\over{#2}}}
\noindent
Find the $y$ coordinate of 
the centroid of the shaded region.
The $x$ coordinate is $0$ by symmetry.
\vtop to 0pt{\vskip 3.75in\LoadPDF{region06.pdf}{0.5}{1in}{1in}\vss}

\vtop to 0pt{\vskip 2.75in \noindent
Find the volume of the solid of revolution obtained by
rotating the shaded region about the line $x=2$.\vss}
\vskip 2.95in
\dquiz
To find the centroid, first find the area:
$$\int_{-1}^1 (3-2x^2) - x^2 dx = \int^1_{-1}3 - 3 x^2 dx = 3x - x^3 \big\vert_{-1}^1 = 
3-\frac13 -( -3 - \frac{-1}{3}) = \frac{5}{3} + \frac{5}{3} = \frac{10}{3}\ .$$
The moment about the $x${--}axis is given by:
\par\noindent
$$\eqalign{%
\int_{-1}^1 \frac{(3-2x^2) + x^2}{2}\ &\bigl((3-2x^2) - x^2\bigr) dx = 
\int_{-1}^1 \frac{ (3 - 2x^2)^2 -(x^2)^2}{2} \ dx\cr
 =\ & \frac12\int_{-1}^1 9 -12 x^2 + 4x^4 - x^4\ dx\cr
 =\ & \frac12\int_{-1}^1 9 -12 x^2 + 3x^4\ dx 
 = \frac12\ \bigl(9x -4x^3 +\frac{3x^5}{5}\bigr) \Big \vert^1_{-1}\cr
 =\ & \frac12 \bigl( (5+ \frac35) - ( -5 - \frac{3}{5}) \bigr) = 5 + \frac35 = \frac{28}{5}
 \cr}$$
 The $y$ coordinate of the center of mass is at $\displaystyle\frac{\frac{28}{5}}{\frac{10}{3}} = 
 \frac{3\cdot 28}{5\cdot 10} = \frac{42}{25}$.
 
 To do the second part, compute the distance from the center of mass at $\bigl(0,\frac{42}{25}\bigr)$
 to the line $x=2$: this distance is $2$.
 Then Pappus says that the volume is $2\pi \cdot 2 \cdot \frac{10}{3} = \frac{40\pi}{3}$.
\equiz
%######################
\stp
%:7
\centerline{\bf Math.~126 Quiz \#7}
\centerline{March  23, 2004}
\vskip10pt
\noindent
\def\frac#1#2{{{#1}\over{#2}}}
\noindent
Find the solution of the differential equation with initial value
$$y\ \frac{d\>z}{d\>y} +\ 2\ z = y^3$$
subject to the initial condition $z(1)=0$.
\dquiz
This equation is linear but not in standard form: the standard form is
$$\frac{d\>z}{d\>y} +\ \frac{2}{y}\ z = y^2\ .$$
Hence $\int P dy = \int \frac{2}{y} dy = 2\ln\vert y\vert +C$.
For an integrating factor, take $e^{\int P dy}= e^{2\ln\vert y\vert} = \vert y\vert^2$ and as usual
for integrating factors one may use $y^2$.
Hence $y^2 z = \int y^2 y^2 dy = \frac{y^5}{5} + C$ so the general solution is
$$z = \frac{y^3}{5} + \frac{C}{y^2}$$
To get the initial value, observe
$$0 = z(1) = \frac{1^3}{5} + \frac{C}{1^2}$$
so $C = -\frac{1}{5}$ and the solution is
$$z =  \frac{y^3}{5} - \frac{1}{5y^2}$$
or if you prefer $z = \frac{y^5-1}{5y^2}$.
\equiz
%######################
\stp
%:8
\centerline{\bf Math.~126 Quiz \#8}
\centerline{March  30, 2004}
\vskip10pt
\noindent
\def\frac#1#2{{{#1}\over{#2}}}
\noindent
Find the area of the region inside the polar curve
$r = 4\cos\theta$ and to the right of the line
$r=\sec \theta$.

\vskip 8true cm
\hskip -2in\vtop to 0pt{\LoadPDF{regionquiz8.pdf}{.5}{2in}{2in}\vss}
\vskip 20pt
\dquiz
The formula is
$$Area = \frac12\ \int_{\alpha_0}^{\alpha_1} (4\cos\theta)^2 - \sec^2\theta\ d\theta$$
where $\alpha_0$ and $\alpha_1$ are the angles where the line and the circle intersect.
These occur when $4\cos\theta = \sec\theta$, or when $4\cos^2\theta = 1$ or
$\cos\theta = \pm\frac{1}{2}$.
From the graph, the $\alpha_i$ are the angles where $\cos\theta = \frac12$ so
$\alpha_1 = \frac{\pi}{3}$ and $\alpha_0 = -\frac{\pi}{3}$.
$$\eqalign{%
Area = &\frac12\ \int_{-\frac{\pi}{3}}^{\frac{\pi}{3}} \ 16\cos^2\theta - \sec^2\theta\ d\theta = 
 \frac12\ \int_{-\frac{\pi}{3}}^{\frac{\pi}{3}} \ 8 - 8\cos(2\theta) - \sec^2\theta\ d\theta\cr
= & 4\>\theta + 2\sin(2\theta) - \frac12 \tan\theta \Big\vert^{\frac{\pi}{3}}_{-\frac{\pi}{3}} = 
\frac{8\pi}{3} + 2(\frac{1}{2} - \frac{-1}{2}) -\frac{1}{2}\bigl(\sqrt{3} - (-\sqrt{3})\bigr)\cr =& 
\frac{8\pi}{3} + 2 -\sqrt{3}
\cr
}$$
\equiz
%######################
\stp
%: 9
\centerline{\bf Math.~126 Quiz \#9}
\centerline{April  6, 2004}
\vskip10pt
\noindent
\def\frac#1#2{{{#1}\over{#2}}}
\def\sL#1{\displaystyle\lim_{#1\to\infty}}
\def\isum#1{{\displaystyle{\sum_{n={#1}}^{\infty}}}\>}

\noindent
Tell why the following series diverges or converges.
If it converges, sum it.
\vskip4pt
$\displaystyle \sum_{n=0}^\infty \frac{2^n + 3^{2n}}{3^{3n+1}}$
\dquiz
Write $\displaystyle \sum_{n=0}^\infty \frac{2^n + 3^{2n}}{3^{3n+1}} =
\sum_{n=0}^\infty \frac{2^n}{3^{3n+1}} +\sum_{n=0}^\infty \frac{3^{2n}}{3^{3n+1}} = 
\sum_{n=0}^\infty \frac{2^n}{3^{3n+1}} +\sum_{n=0}^\infty \frac{1}{3^{n+1}}$.

Each of these two series is geometric:
$\displaystyle\sum_{n=0}^\infty \frac{2^n}{3^{3n+1}}$ has $r = \frac{2}{27}$ and $a = \frac13$,
so $\displaystyle\sum_{n=0}^\infty \frac{2^n}{3^{3n+1}} = \frac{\smfrac13}{1 - \smfrac{2}{27}} =
\frac{\smfrac13}{\smfrac{25}{27}} = \frac{9}{25}$
and $\displaystyle\sum_{n=0}^\infty \frac{1}{3^{n+1}}$ has $r=\frac13$ and $a=\frac13$, so
$\displaystyle\sum_{n=0}^\infty \frac{1}{3^{n+1}} = \frac{\smfrac13}{1 - \smfrac13}
=  \frac{\smfrac13}{\smfrac23} = \frac12$.

Hence 
$\displaystyle \sum_{n=0}^\infty \frac{2^n + 3^{2n}}{3^{3n+1}} = \frac{9}{25} + \frac{1}{2} = 
\frac{18 + 25}{50} = \frac{43}{50}$.
\equiz
%######################
\stp
%:10
\centerline{\bf Math.~126 Quiz \#10}
\centerline{April  13, 2004}
\vskip10pt
\noindent
\def\frac#1#2{{{#1}\over{#2}}}
\def\sL#1{\displaystyle\lim_{#1\to\infty}}
\def\isum#1{{\displaystyle{\sum_{n={#1}}^{\infty}}}\>}

\noindent
Tell why each of the following series diverges or converges.
\vskip4pt
\item{1.}
$\displaystyle \sum_{n=1}^\infty \frac{1}{n}$
\item{2.}
$\displaystyle \sum_{n=1}^\infty \frac{1}{n^2}$
\item{3.}
$\displaystyle \sum_{n=1}^\infty \frac{\ln n}{n^2}$
\dquiz
\item{\#1:} This is the harmonic series, or the $p${--}series with $p=1$.
Hence it diverges since $1=p \leq 1$
\item{\#2:} This is the $p${--}series with $p=2$.
Hence it converges since $2=p > 1$
\item{\#3:} Since $\ln n$ grows more slowly than powers of $n$, {\sl eventually\/}
$\ln n < \sqrt{n\;}$ so $\displaystyle\frac{\ln n}{n^2} < \frac{1}{n^{\smfrac{3}{2}}}$.
Now $\displaystyle \sum_{n=1}^\infty \frac{1}{n{\smfrac{3}{2}}}$ is the $p$ series
with $\smfrac32 = p > 1$ so it converges.
By the Comparison Test $\displaystyle \sum_{n=1}^\infty \frac{\ln n}{n^2}$
also converges.

\item{}To actually check $\ln n < \sqrt{n\;}$ eventually, consider the function
$f(x) = \sqrt{x\;} - \ln x$. Compute 
$\displaystyle f^\prime(x)= \frac{1}{2\sqrt{x\;}} - \frac{1}{x} = \frac{\sqrt{x\;} - 2}{2x}$
and this is positive for $x>4$.
Check $f(4)= 2 -\ln 4$ and $e^2 > (2.5)^2 >6>4$ so $\ln 4 < 2$ and $f(4)>0$.
Therefore $f(x)>0$ for all $x\geq 4$.

\equiz
%######################
\stp
%:11
\centerline{\bf Math.~126 Quiz \#11}
\centerline{April  27, 2004}
\vskip10pt
\noindent
\def\frac#1#2{{{#1}\over{#2}}}
\def\sL#1{\displaystyle\lim_{#1\to\infty}}
\def\isum#1{{\displaystyle{\sum_{n={#1}}^{\infty}}}\>}

\noindent
Find the interval of convergence for the power series

$$\sum_{n=1}^{\infty}\ \frac{x^n}{\ln(n+1)}$$
\dquiz
First compute the radius of convergence.
Via the ratio test:
$\displaystyle \left\vert \frac{\smfrac{x^{n+1}}{\ln(n+2)}}{\smfrac{x^n}{\ln(n+1)}}\right\vert =
\frac{\ln(n+1)}{\ln(n+2)} \vert x \vert$.
By l'H\^{o}pital's Rule $\displaystyle\lim_{n\to\infty}\ \frac{\ln(n+1)}{\ln(n+2)} =
\lim_{n\to\infty}\ \frac{\smfrac{1}{n+1}}{\smfrac{1}{n+2}} = \lim_{n\to\infty}\ \frac{n+2}{n+1} = 1$
and hence the radius of convergence is $1$.

Via the root test $\displaystyle \left\vert \root n \of{\frac{x^n}{\ln(n+1)}\;}\right\vert = 
\frac{\vert x\vert} {\bigl(\ln(n+1)\bigr)^{\smfrac{1}{n}}}$.
This involves computing 
$$\displaystyle\lim_{n\to\infty}\ \bigl(\ln(n+1)\bigr)^{\smfrac{1}{n}}$$
which is an indeterminate exponential.
Hence compute 
$$\lim_{x\to\infty}\ \frac{1}{x}\ \ln\bigl(\ln(x+1)\bigr) = 
\lim_{x\to\infty}\ \frac{\frac{\frac{1}{x+1}}{\ln(x+1)}}{1} = 
\lim_{x\to\infty}\ \frac{1}{(x+1)\ln(x+1)}\ .$$

$\lim_{x\to\infty}\ (x+1)\ln(x+1)$ is another indeterminate form, so
$$\lim_{x\to\infty}\ (x+1)\ln(x+1) = \lim_{x\to\infty}\ \frac{\ln(x+1)}{\smfrac{1}{x+1}} =
\lim_{x\to\infty}\ \frac{\smfrac{1}{x+1}}{\smfrac{-1}{(x+1)^2}} = 
\lim_{x\to\infty}\ -(x+1) = -\infty\ .
$$
It follows that $\displaystyle\lim_{x\to\infty}\ \frac{1}{x}\ \ln\bigl(\ln(x+1)\bigr) = \frac{1}{-\infty} = 0$
and hence $\displaystyle\lim_{n\to\infty}\ \bigl(\ln(n+1)\bigr)^{\smfrac{1}{n}} = e^0 = 1$.
Again it follows that the radius of convergence is $1$.

To determine the interval of convergence we need to examine the series
$\displaystyle\sum_{n=1}^{\infty}\ \frac{1}{\ln(n+1)}$ and
$\displaystyle\sum_{n=1}^{\infty}\ \frac{(-1)^n}{\ln(n+1)}$.
Since $\ln(n+1)$ is eventually less than $n$, and since $\displaystyle\sum_{n=1}^{\infty}\ \frac{1}{n}$
is the harmonic series which diverges, the Comparison Test shows 
$\displaystyle\sum_{n=1}^{\infty}\ \frac{1}{\ln(n+1)}$ diverges.

The series $\displaystyle\sum_{n=1}^{\infty}\ \frac{(-1)^n}{\ln(n+1)}$ is alternating and the terms go to
$0$: i.e. $\displaystyle\lim_{n\to\infty}\ \frac{1}{\ln(n+1)} = 0$.
If $f(x) = \frac{1}{\ln(x+1)}$, $f^\prime(x) = \frac{-1}{\bigl(\ln(x+1)\bigr)^2}\ \frac{1}{x+1}$
which is $< 0 $ for $x>0$.
By the Alternating Series Test $\displaystyle\sum_{n=1}^{\infty}\ \frac{(-1)^n}{\ln(n+1)}$
converges.

Hence the interval of convergence is $[-1,1)$.
\equiz

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