The ratio test says that the series [tex]\sum a_n[/tex] converges if the absolute value of the ratio between consecutive terms approaches a number [tex]0\le L<1[/tex], and diverges otherwise:
[tex]\displaystyle\lim_{n\to\infty}\left|\frac{a_{n+1}}{a_n}\right|<1[/tex]
You have
[tex]\displaystyle\lim_{n\to\infty}\left|\frac{\frac{2^{n+1}}{x^{n+1}}}{\frac{2^n}{x^n}}\right|=\lim_{n\to\infty}\left|\frac{2^{n+1}}{2^n}\frac{x^n}{x^{n+1}}\right|=\lim_{n\to\infty}\left|\frac2x\right|=\frac2{|x|}\lim_{n\to\infty}1=\frac2{|x|}[/tex]
Convergence requires that this limit be smaller than 1, so this happens when [tex]\dfrac2{|x|}<1\implies|x|>2[/tex]. (This agrees with what you already know about geometric series; [tex]\dfrac{2^n}{x^n}=\left(\dfrac2x\right)^n[/tex] only converges if [tex]\left|\dfrac2x\right|<1[/tex].)
The ratio test also works with the second problem.
[tex]\displaystyle\lim_{n\to\infty}\left|\frac{\frac{\sin^{n+1}x}{3^{n+1}}}{\frac{\sin^nx}{3^n}}\right|=\lim_{n\to\infty}\left|\frac{\sin x}3\right|=\dfrac{|\sin x|}3<1[/tex]
[tex]\implies |\sin x|<3[/tex]
But recall that for all [tex]x[/tex], you have [tex]|\sin x|\le1[/tex], so the above condition is always met. This means the series converges everywhere.
These first two are also good candidates for using the root test.
For the last two, try the integral test.
[tex]\dfrac1{n(\ln n)^p}[/tex] will be positive for all [tex]p[/tex] so long as [tex]n\ge3[/tex] ([tex]\ln2[/tex] is negative, and so the first term in the series will be negative for any odd [tex]p[/tex]; but we can just omit this term from the series because any one term is still a finite number). So, we compare to the integral,
[tex]\displaystyle\int_3^\infty\frac{\mathrm dx}{x(\ln x)^p}\le\sum_{n\ge3}\frac1{n(\ln n)^p}[/tex]
Let [tex]y=\ln x[/tex], so that [tex]\mathrm dy=\dfrac{\mathrm dx}x[/tex]. Then
[tex]\displaystyle\int_3^\infty\frac{\mathrm dx}{x(\ln x)^p}=\int_{\ln3}^\infty\frac{\mathrm dy}{y^p}[/tex]
[tex]=\displaystyle\lim_{t\to\infty}\int_{\ln3}^t\frac{\mathrm dy}{y^p}[/tex]
[tex]=\displaystyle\lim_{t\to\infty}\frac{y^{1-p}}{1-p}\bigg|_{y=\ln3}^{y=t}[/tex]
[tex]=\displaystyle\lim_{t\to\infty}\frac{t^{1-p}}{1-p}-\frac{(\ln3)^{1-p}}{1-p}[/tex]
Note that we immediately know [tex]p\neq1[/tex]. You have
[tex]\displaystyle\lim_{t\to\infty}\frac{t^{1-p}}{1-p}=\begin{cases}\infty&\text{for }1-p\ge0\\0&\text{for }1-p<0\end{cases}[/tex]
so the integral converges only for [tex]1-p\le0\implies p>1[/tex].
For the last one, omit the first two terms ([tex]n=1\implies\dfrac{\ln n}{n^p}=0[/tex] and [tex]n=2\implies\dfrac{\ln n}{n^p}<0[/tex]). The remaining sequence is positive and decreasing, which follows from the fact that [tex]\ln n<n<n^p[/tex]. Then compare to the definite integral,
[tex]\displaystyle\int_3^\infty\frac{\ln x}{x^p}\,\mathrm dx\le\sum_{n\ge3}\frac{\ln n}{n^p}[/tex]
Integrate by parts, setting [tex]u=\ln x[/tex] and [tex]\mathrm dv=x^{-p}\,\mathrm dx[/tex], so that [tex]\mathrm du=\dfrac{\mathrm dx}x[/tex] and [tex]v=\dfrac{x^{1-p}}{1-p}[/tex]. Then
[tex]\displaystyle\int_3^\infty\frac{\ln x}{x^p}\,\mathrm dx=\frac{x^{1-p}\ln x}{1-p}\bigg|_{x=3}^{x\to\infty}-\frac1{1-p}\int_3^\infty x^{2-p}\,\mathrm dx[/tex]
[tex]=\displaystyle\lim_{t\to\infty}\frac{t^{1-p}\ln t}{1-p}-\frac{(\ln3)^{1-p}\ln(\ln3)}{1-p}-\frac{x^{1-p}}{(1-p)^2}\bigg|_{x=\ln3}^{x\to\infty}[/tex]
[tex]=\displaystyle\lim_{t\to\infty}\left(\frac{t^{1-p}\ln t}{1-p}-\frac{t^{1-p}}{(1-p)^2}\right)+\frac{(\ln3)^{1-p}}{(1-p)^2}-\frac{(\ln3)^{1-p}\ln(\ln3)}{1-p}[/tex]
[tex]=\displaystyle\frac1{1-p}\lim_{t\to\infty}\left(\frac{t^{1-p}\ln t}{1-p}-\frac{t^{1-p}}{(1-p)^2}\right)+\frac{(\ln3)^{1-p}}{(1-p)^2}-\frac{(\ln3)^{1-p}\ln(\ln3)}{1-p}[/tex]
Again, we can't have [tex]p=1[/tex]. The integral will converge if the limit converges to a finite number.
[tex]\displaystyle\lim_{t\to\infty}\left(\frac{t^{1-p}\ln t}{1-p}-\frac{t^{1-p}}{(1-p)^2}\right)=\frac1{(1-p)^2}\lim_{t\to\infty}t^{1-p}((p-1)\ln t-1)[/tex]
When [tex]p<1[/tex], the limit diverges to [tex]\infty[/tex]. When [tex]p>1[/tex], evaluating the limit directly gives an indeterminate form [tex]0\times\infty[/tex]. We rewrite this as
[tex]\displaystyle\lim_{t\to\infty}\frac{(p-1)\ln t-1}{t^{p-1}}[/tex]
and using the fact that [tex]\ln t<t<t^p[/tex] for large enough [tex]t[/tex] whenever [tex]p>1[/tex], it follows that the limit exists by the squeeze theorem. So L'Hopital's rule applies here, and the limit is equal to
[tex]\displaystyle\lim_{t\to\infty}\frac{(p-1)\ln t-1}{t^{p-1}}=\lim_{t\to\infty}\frac{\frac{p-1}t}{(p-1)t^{p-2}}=\lim_{t\to\infty}t^{p-1}=0[/tex]
All this to say the last series converges for [tex]p>1[/tex].
