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Question

In this task, determine the convergence or divergence of the series.

n=21n(lnn)3\displaystyle\sum_{n=2}^{\infty} \frac{1}{n(\ln n)^3}

Solution

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Answered 2 years ago
Answered 2 years ago
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Let f(x)=1x(lnx)3f(x)=\dfrac{1}{x(\ln{x})^3} then f(x)f(x) is positive and continuous for x2x\geq 2. Consider

f(x)=(lnx)33(lnx)2x2(lnx)6=lnx3x2(lnx)4lt;0for x2\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 ff is decreasing for x2x\geq 2. So we can apply integral's test.

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