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Question

# In this task, determine the convergence or divergence of the series.$\displaystyle\sum_{n=2}^{\infty} \frac{1}{n(\ln n)^3}$

Solution

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Let $f(x)=\dfrac{1}{x(\ln{x})^3}$ then $f(x)$ is positive and continuous for $x\geq 2$. Consider

\begin{aligned}f'(x)&=\dfrac{(\ln{x})^3-3(\ln{x})^2}{x^2(\ln{x})^6}\\&=\dfrac{-\ln{x}-3}{x^2(\ln{x})^4}\\&<0\quad \text{for }x\geq 2\end{aligned}

which implies $f$ is decreasing for $x\geq 2$. So we can apply integral's test.

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