Integral Test — Question 7

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Question 7

Consider ∑n=1∞1n2+4.\sum_{n=1}^{\infty}\frac1{n^2+4}.

  1. Verify positivity, continuity, and decreasing behavior for f(x)=1/(x2+4)f(x)=1/(x^2+4).

  2. Use the Integral Test to classify the series, showing the arctangent antiderivative and improper limit.

  3. Write explicit upper and lower bounds for the remainder after NN terms.

Your remainder bounds should contain NN and inverse tangent functions.

Original worksheet page 1: question and worked solution for 4-6-007
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Question 7 – Solution

Step 1: Verify the hypotheses.

Define f(x)=1/(x2+4)f(x)=1/(x^2+4) for x≥1x\ge1. It is continuous and positive, and f′(x)=−2x(x2+4)2<0,f'(x)=-\frac{2x}{(x^2+4)^2}<0, so it decreases.

Step 2: Test the improper integral.

∫1∞dxx2+4=12[arctan(x2)]1∞=12(π2−arctan12)<∞.\int_1^{\infty}\frac{dx}{x^2+4} =\frac12\left[\arctan\left(\frac x2\right)\right]_1^{\infty} =\frac12\left(\frac\pi2-\arctan\frac12\right)<\infty. Therefore the series converges.

Step 3: Estimate the remainder.

For A>0A>0, ∫A∞dxx2+4=12(π2−arctanA2).\int_A^{\infty}\frac{dx}{x^2+4} =\frac12\left(\frac\pi2-\arctan\frac A2\right). Consequently, 12(π2−arctanN+12)≤RN≤12(π2−arctanN2).\frac12\left(\frac\pi2-\arctan\frac{N+1}{2}\right) \le R_N\le \frac12\left(\frac\pi2-\arctan\frac N2\right).

Original worksheet page 2: question and worked solution for 4-6-007

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