Convergence of these series $$\sum_{n=1}^\infty \frac{2nx^{n}}{(n+1)^{2}3^{n}} \tag{1}$$
$$\sum_{n=1}^\infty x^{n}\tan\frac{x}{4^{n}}\tag{2}$$
Is there any good article that describes an equivalents like if $$ \lim_{n\to\infty} \sin\frac{2\pi}{3^{n}} \sim \frac{2\pi}{3^{n}}\tag{3}$$
(am I right about $(3)$?)
 A: $\lim_{n\to\infty} \sin\frac{2\pi}{3^{n}} \sim \frac{2\pi}{3^{n}}\tag{3}$ Is correct since $\lim_{x\to 0 }  \dfrac{sinx}{x}=1$ and since sinus is continuous this holds for any sequence 
 ${x_n\to 0}$
A: The assertion
$$ \lim_{x\to\infty} \sin\frac{2\pi}{3^{n}} \sim \frac{2\pi}{3^{n}}$$
is perhaps not precise enough for our needs. But it can be made precise, in a form useful for Ratio Test calculations. 
We have 
$$\lim_{t\to 0}\frac{\sin t}{t}=1.$$
We can replace $\sin\left(\frac{2\pi}{3^n}\right)$ by 
$$\frac{\sin\left(\frac{2\pi}{3^n}\right)}{\frac{2\pi}{3^n}}\frac{2\pi}{3^n}.$$
The first term goes safely to $1$, while $\frac{2\pi}{3^n}$ "plays nicely" with the Ratio Test. 
We can do precisely the same thing with $\tan\left(\frac{x}{4^n}\right)$ for $x\ne 0$.  Since 
$$\lim_{t\to 0}\frac{\tan t}{t}=1,$$ 
use
$$\frac{\tan\left(\frac{x}{4^n}\right)}{\frac{x}{4^n}}\frac{x}{4^n}$$
in the Ratio Test limit calculation. The front part has limit $1$, so it ford not affect the Ratio Test limit. 
Remark: More informally, we can simply replace $\tan(x/4^n)$ by $x/4^n$. However, that is likely to be considered too informal in a homework problem when an important  focus of the course is careful justification. Later, when one is doing "real" work, we can be more casual, for we know that formal detail could be supplied. 
In answer to an earlier comment of yours, $\cos x$ is nearly $1$ if $x$ is close to $0$. But it can be useful to know that since the first $2$ terms of the power series of $\cos x$ are $1-\frac{x^2}{2}$, near $0$ the function $1-\cos x$ behaves like $\frac{x^2}{2}$. 
