Determine the convergence or divergence of the series.sum_{n=3}^inftyfrac{1}{n(ln n)[ln(ln n)]^4}

A series, $\sum _{n=m}^{\mathrm{\infty }}f(n)$ converges if the value of the integral, ${\int }_m^{\mathrm{\infty }}f(n)$ is finite and diverges if it is infinite.
For the series, $\sum _{n=3}^{\mathrm{\infty }}\frac{1}{n(\ln n)[\ln (\ln n){]}^4}$, the integral is ${\int }_3^{\mathrm{\infty }}\frac{1}{x(\ln x)[\ln (\ln x){]}^4}dx$
Let $\ln (\ln x)=y,\frac{1}{x\ln x}dx=dy$
At $x=3,y=\ln (\ln 3)$
at $x=\mathrm{\infty },y=\mathrm{\infty }$
So, the integral becomes:
${\int }_3^{\mathrm{\infty }}\frac{1}{x(\ln x)[\ln (\ln x){]}^4}dx={\int }_{\ln (\ln 3)}^{\mathrm{\infty }}\frac{1}{x(\ln x)(y{)}^4}x(\ln x)dy$
$={\int }_{\ln (\ln 3)}^{\mathrm{\infty }}\frac{1}{y^4}dy$
$=-\frac{1}{5y^5}{|}_{\ln (\ln 3)}^{\mathrm{\infty }}$
$=-\frac{1}{5}(\frac{1}{\mathrm{\infty }}-\frac{1}{(\ln (\ln 3){)}^5})$
$=\frac{1}{5(\ln (\ln 3){)}^5}$
Since, the value of the integral is finite. Thus, the sum of the series converges.