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

Determine the convergence or divergence of the series.
$\sum _{n=3}^{\mathrm{\infty }}\frac{1}{n\left(\mathrm{ln}n\right)\left[\mathrm{ln}\left(\mathrm{ln}n\right){\right]}^{4}}$

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