Is this a correct proof of the divergence of a series? I wanted to show if
$$ \sum_{n=2}^\infty \frac{\sqrt{n}}{3^n\ln(n)} $$
converges or diverges. By the ratio test, I could solve for $a_n=\frac{\sqrt{n}}{3^n\ln(n)}$ the limit
$$\lim_{n\to\infty} \frac{a_{n+1}}{a_n}$$
and see what the result tells me.
$Proof:$
$i)$ Let $f(x)=\frac{\sqrt{x}}{3^x\ln(x)}$. Then, because $f(n)=a_n$ and $f(n+1)=a_{n+1}$, 
$$\lim_{n\to\infty} \frac{a_{n+1}}{a_n}=\lim_{x\to\infty} \frac{f(x+1)}{f(x)}$$
$ii$) $$\lim_{x\to\infty} \frac{f(x+1)}{f(x)}= \lim_{x\to\infty} \frac{\sqrt{x+1}}{3^{x+1}\ln(x+1)}*\frac{3^x \ln(x)}{\sqrt{x}}=\frac{1}{3}\lim_{x\to\infty} \frac{\sqrt{x+1}}{\sqrt{x}} *\frac{\ln(x)}{\ln(x+1)} $$
$$= \frac{1}{3}\lim_{x\to\infty} \sqrt{\frac{x+1}{x}}*\frac{x+1}{x}$$
$iii)$ It is not hard to show that $\lim_{x\to\infty} \sqrt{\frac{x+1}{x}}$ and $\lim_{x\to\infty} \frac{x+1}{x}$ both exist and are equal to one. Then
$$= \frac{1}{3}\lim_{x\to\infty} \sqrt{\frac{x+1}{x}}*\frac{x+1}{x} = \frac{1}{3}(\lim_{x\to\infty} \sqrt{\frac{x+1}{x}}*\lim_{x\to\infty} \frac{x+1}{x})=\frac{1}{3}$$
Because of $i$ we conclude that $$\lim_{n\to\infty} \frac{a_{n+1}}{a_n} = \frac{1}{3}<1$$
$iiii$) Because the ratio is less than one, then the series converges.

It's correct. It can also be done without switching to functional limits and using L'Hôpital's rule:
$$\frac{a_{n+1}}{a_n}=\frac 13\sqrt{1+\frac 1n}\frac{\ln n}{\ln(n+1)}\to \frac 13\cdot 1\cdot 1=\frac 13<1 $$
To handle the fraction with the logarithms
$$\frac{\ln(n+1)}{\ln n}=\frac{\ln n+\ln\left(1+\frac 1n\right)}{\ln n}=1+\ln\left(1+\frac 1n\right)\cdot\frac{1}{\ln n}\to1+0\cdot 0=1 $$
so the reciprocal $\ln(n)/\ln(n+1)$ also has limit $1$.
