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\newp 12. 
 This is a geometric series with the first term being 
$e^{2-1}=e$, and the ratio
of terms being $\frac{e^{2-(k+1)}}{e^{2-k}}=\frac{1}{e}$.  The sum is 
therefore $\frac{e}{1-\frac{1}{e}}= \frac{e^2}{e-1}$.
\endp

\newp 13.  The ratio test may be used to determine 
the radius of convergence.
$$\lim_{k\rightarrow \infty}\Big\vert {a_{k+1}\over 
a_k}\Big\vert=\lim_{k\rightarrow \infty}{\vert
x\vert^{k+1}\over k+1}{k\over\vert x\vert^k}=\vert x\vert$$
Hence the radius of convergence is one.

The left endpoint of the interval of convergence is $-1$, and 
$\displaystyle\sum_{k=1}^\infty {(-1)^k\over k}$ 
converges by the Alternating Series Test.

The right endpoint of the interval of convergence is $+1$, and 
$\displaystyle\sum_{k=1}^\infty {1\over k}$ diverges 
either because it is the harmonic series, or 
because it is a $p$-series
with $p=1$.

So the interval of convergence is $[-1,1)$.
\endp

\newp 14.
To use the Comparison Test one has to find a series to which to compare
the given one.
Look at the $k$th term, $\frac{1}{2^k+k}$: this can easily be
compared to 
\itemitem{1)} $\frac{1}{2^k+k} < \frac{1}{2^k}$
\itemitem{2)} $\frac{1}{2^k+k} < \frac{1}{k}$

If you guessed 2), then $\displaystyle\sum_{k=1}^\infty\frac{1}{k}$
is the harmonic series (or $p${--}series with $p=1$) and hence it
diverges.
But this says nothing about the original series.

Try 1): $\displaystyle\sum_{k=1}^\infty\frac{1}{2^k}$ is a geometric
series with $r=\frac{1}{2}<1$ and so it converges and now the
Comparison Test says $\displaystyle\sum_{k=1}^\infty\frac{1}{2^k+k}$
converges.

It is also possible, although harder, to compare to 
$\displaystyle\sum_{k=1}^\infty\frac{1}{k^2}$.
This is a $p${--}series with $p=2$ and so converges.
Hence we would like to show
$\frac{1}{2^k+k} < \frac{1}{k^2}$, at least for large $k$.
Equivaelently,
$k^2 < 2^k+k$.
To do this, let $f(x)=2^x+x-x^2$.
As $x\to\infty$, $f \to \infty$ so $f(x)>0$
for all large enough $x$.
In other words, $x^2< 2^x+x$ for all large $x$. 
\endp


\newp 15.
Two curves have two points of intersection. First point is the
origin: (0,0). In order to find second point we need to equate two
expressions $ 2 \sqrt{3} \cos \theta = 2 \sin \theta$. This
results in $\tan \theta = \sqrt{3}$. Lastly, we have
$\theta = \frac{\pi}{3}$. The area of the domain bounded by two
curves is given by the following formula 
$$\frac{1}{2}
\int_0^{\smfrac{\pi}{3}} (2 \sin \theta)^2 \; d \theta + \frac{1}{2}
\int_{\smfrac{\pi}{3}}^{\smfrac{\pi}{2}} (2 \sqrt{3} \cos \theta)^2 \;
d \theta = 2 \int_0^{\smfrac{\pi}{3}}  \sin^2 \theta \; d \theta + 6
\int_{\smfrac{\pi}{3}}^{\smfrac{\pi}{2}}  \cos^2 \theta \; d \theta.
$$ 
After using the following substitutions: $\sin^2 \theta =
\frac{1}{2} (1 - \cos 2 \theta)$ and $\cos^2 \theta = \frac{1}{2}
(1 + \cos 2 \theta)$ previous formula yields that 
$$
\int_0^{\smfrac{\pi}{3}} (1- \cos 2 \theta) \; d \theta + 3
\int_{\smfrac{\pi}{3}}^{\smfrac{\pi}{2}} (1+ \cos 2\theta) \; d \theta
= \frac{5 \pi}{6} -\sqrt{3}\  .$$

\endp


\end