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Determine if the series, a1 + a2 + a3 + a4 +… converges or diverges, when:
1. an = n/(n2 + 1)
Answer
1. Diverges
Solution
1. f(x)= x/(x2 + 1) is positive, continuous, and decreasing because f'(x)= (1 – x2)/(1 + x2)2 is negative for x > 1. Substituting u= x2 + 1 so du= 2x dx and x dx = (1/2)du, we see ∫x/(x2 + 1) dx = (1/2) ∫du/u = (1/2)ln(u) = (1/2)ln(x2 + 1) This diverges to ∞ as x → ∞, so the series diverges by the Integral Test.
2.an= n2/(n3 + 1)
Answer
1. Diverges
Solution
1. f(x)= x2/(x3 + 1) is positive, continuous. Now f'(x)= x(2 – x3)/(1 + x3)2 is negative for x > 21/3≈ 1.2599, so we can apply the Integral test only to the series starting with n= 2. This is fine: the n = 1 term cannot affect the convergence of the series. Substituting u= x3 + 1, so du = 3x2 dx and x2 dx = (1/3)du, we see ∫x2/(x3 + 1) dx = (1/3) ∫du/u = (1/3) ln(u) = (1/3) ln(x3 + 1) This diverges to ∞ as x → ∞, so the series diverges by the Integral Test.
3. an= n/(n2 + 1)2
Answer
1. Converges
Solution
1. f(x)= x/(x2 + 1)2 is positive, continuous, and decreasing because f'(x)= (1 – 3x2)/(1 + x2)3is negative for x > 1. Substituting u = x2 + 1 so du = 2x dx and x dx = (1/2)du, we see: ∫x/(x2 + 1)2 dx = (1/2)∫ du/u2 = -(1/2)(1/u) = -(1/2)(1/(x2 + 1)) This converges, so the series converges by the Integral Test.
4. an=1/(n ln(n)), starting with n = 2
Answer
1. Diverges
Solution
1. f(x)= 1/(x ln(x)) is positive, continuous, and decreasing because f'(x)= -(1 + ln(x))/(x2ln2(x)) is negative for > 1. Substituting u= ln(x) so du = (1/x)dx, we see ∫1/(x ln(x)) dx = ∫ du/u = ln(u) = ln(ln(x)) This diverges to ∞ as x → ∞, so the series diverges by the Integral Test.