Strategy for Series — Question 2

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Question 2

Determine whether ∑n=2∞1nlog⁡n\displaystyle\sum_{n=2}^{\infty}\frac1{n\log n} converges or diverges. Explain why the Integral Test is especially well suited to this expression.

Original worksheet page 1: question and worked solution for 4-12-002
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Question 2 – Solution

Step 1: Verify the hypotheses.

Let f(x)=1/(xlog⁡x)f(x)=1/(x\log x) for x≥2x\ge2. It is positive and continuous. Also, f′(x)=−log⁡x+1x2(log⁡x)2<0,f'(x)=-\frac{\log x+1}{x^2(\log x)^2}<0, so ff is decreasing.

Step 2: Evaluate the improper integral.

The integrand contains the differential dx/x=d(log⁡x)dx/x=d(\log x), suggesting u=log⁡xu=\log x: ∫2bdxxlog⁡x=∫log⁡2log⁡bduu=log⁡(log⁡b)−log⁡(log⁡2).\int_2^b\frac{dx}{x\log x}=\int_{\log2}^{\log b}\frac{du}{u}=\log(\log b)-\log(\log2). As b→∞b\to\infty, this tends to ∞\infty.

Conclusion.

The series diverges by the Integral Test. The substitution exposes the antiderivative log⁡(log⁡x)\log(\log x) immediately, which is why this test is more natural than a ratio or root test here.

Original worksheet page 2: question and worked solution for 4-12-002

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