Exercises to see if certain series converge. 
*

*Does the series $\sum \frac{2^k+1}{3^k}$ converge?  

*Does the series $\sum (-1)^k\frac{1}{k^2}$ converge?  

*Does the series $\sum \frac{2^k+2^{2k}}{4^k}$ converge?  

*Does the series $\sum \frac{k!}{(2k)!}$ converge?


For the first question, do we use the comparison test comparing it to $\left(\frac{2}{3}\right)^k$???  
For the second question, this series does converge. To prove this observe that this converges absolutely because
\begin{equation*}
\sum_{k=1}^{\infty} \left|(-1)^k\frac{1}{k^2}\right| = \sum_{k=1}^{\infty} \frac{1}{k^2}
\end{equation*}
converges. To see that this does converge, we can use the integral test.
The function $f:[1,\infty) \mapsto \mathbb{R}$ defined by $f(x) := \frac{1}{x^2}$ is continuous, non-negative and decreasing by examining the derivative $f'(x) = -\frac{2}{x^3} < 0$ for all $x\geq 1$. So we have
\begin{equation*}
\begin{split}
\int_{1}^{\infty} \frac{1}{x^2} \; dx &= \left[-\frac{1}{x}\right]_{1}^{\infty} \\
&= \lim_{a\to\infty} \left(-\frac{1}{a}+1\right) \\
&= 1
\end{split}
\end{equation*}
This integral converges so by the integral test $\sum_{k=1}^{\infty} \frac{1}{k^2}$ converges as well and hence, $\sum_{k=1}^{\infty} (-1)^k\frac{1}{k^2}$ converges.  
For the third question, this series does NOT converge. To show that this series diverges, we just need to show that
\begin{equation*}
\lim_{k\to\infty} \frac{2^k+2^{2k}}{4^k} \neq 0
\end{equation*}
or doesn't exist. Well, for $k\geq 0$,
\begin{equation*}
\frac{2^k+2^{2k}}{4^k} = \frac{2^k+4^k}{4^k} = \left(\frac{1}{2}\right)^k+1.
\end{equation*}
Thus,
\begin{equation*}
\begin{split}
\lim_{k\to\infty} \frac{2^k+2^{2k}}{4^k} &= \lim_{k\to\infty} \left(\frac{1}{2}\right)^k+1 \\
&= 0+1 \\
&= 1 \neq 0.
\end{split}
\end{equation*} 
For the fourth question, if we let $a_k = \frac{k!}{(2k)!}$ then using the ratio test we get
\begin{equation*}
\begin{split}
\lim_{k\to\infty} \left|\frac{a_{k+1}}{a_k}\right| &= \lim_{k\to\infty} \left|\frac{\frac{(k+1)!}{(2k+1)!}}{\frac{k!}{(2k)!}}\right| \\
&= \lim_{k\to\infty} \left|\frac{(k+1)!}{(2k+1)!}\times \frac{(2k)!}{k!}\right| \\
&= \lim_{k\to\infty} \left|\frac{(k+1)\times k\times (k-1)\times\ldots\times (2k)\times (2k-1)\times (2k-2)\times\ldots}{(2k+1)\times 2k\times (2k-1)\times\ldots\times k\times (k-1)\times (k-2)\times\ldots}\right| \\
&= \lim_{k\to\infty} \frac{k+1}{2k+1} \\
&= \lim_{k\to\infty} \frac{1+\frac{1}{k}}{2+\frac{1}{k}} \\
&= \frac{1}{2} < 1.
\end{split}
\end{equation*}
Hence, $\sum_{k=0}^{\infty} \frac{k!}{(2k)!}$ converges.

For 1) use the fact that $\sum (\frac 2 3)^{k}$ and $\sum \frac1 {3^{k}}$ both converge. 
Your answer for 2) and 3) are OK. 
In 4) $a_{k+1}$ is not $\frac {(k+1)!} {(2k+1)!}$. It is $\frac {(k+1)!} {(2k+2)!}$. [You have to replace $k$ by $k+1$ in the formula for $a_k$]. But your approach is good and you will be able to complete it after making this correction. 
