Integral Test — Question 7

Section 11-6 · Calculus pages 2053–2054

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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.

*Question 7 – Solution

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.

∫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.

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).