For the series, \(\sum_{n=3}^\infty\frac{1}{n(\ln n)[\ln(\ln n)]^4}\), the integral is \(\int_3^\infty\frac{1}{x(\ln x)[\ln(\ln x)]^4}dx\)

Let \(\ln(\ln x)=y,\frac{1}{x\ln x}dx=dy\)

At \(x=3,y=\ln(\ln3)\)

at \(x=\infty,y=\infty\)

So, the integral becomes:

\(\int_3^\infty\frac{1}{x(\ln x)[\ln(\ln x)]^4}dx=\int_{\ln(\ln3)}^\infty\frac{1}{x(\ln x)(y)^4}x(\ln x)dy\)

\(=\int_{\ln(\ln3)}^\infty\frac{1}{y^4}dy\)

\(=-\frac{1}{5y^5}|_{\ln(\ln3)}^\infty\)

\(=-\frac15(\frac{1}{\infty}-\frac{1}{(\ln(\ln3))^5})\)

\(=\frac{1}{5(\ln(\ln3))^5}\)

Since, the value of the integral is finite. Thus, the sum of the series converges.