Find value of ∞Σ k=1 2k-3/k^3 to within .01 of its exact value

Question

∫ from n to n+1 2x-3/x^3 dx < 0.1 implies N > 12.8

(a) Check the conditions of any theorem/test you apply. Hint: Conditions of any test just need to be eventually true, i.e., there is an a such that for k >= a the conditions hold.

(b) Apply a theorem/test to find the value of ∞Σ k=1 2k − 3/k^3 to within .01 of its exact value.

Answer 1

Consider using the "integral test" for checking series convergence. As such, above series is convergent if and only if the integral ∫₁^∞ (2x-3)/x^3 dx is convergent. The integral indeed converges to 0.5 so the series is convergent.

Note that: ∑₁^∞ (2k-3)/ k ^3 = ∑₁^a (2k-3)/ k ^3 + ∑_a+1^∞ (2k-3)/ k ^3. If we write this as A= B+C, we can exactly calculate B. So approximating A within 0.01 of its exact value depends on approximating C within within 0.01 of its exact value.

On the other hand: ∑_a^∞ (2k-3)/ k ^3 < ∫_a+1^∞ (2x-3)/x^3 dx = D, because the sum of rectangle areas (corresponding the series) is smaller than the area under the curve (corresponding to integral). By calculating the integral D and simplifying, we obtain that D= (4a-3) /(2a^2) < 1/100 or a > 200. Therefore,

∑₁²⁰⁰ (2k-3)/ k ^3 yields a good estimation of the series A within 0.01 of its exact value or series has to be calculated for at least 200 terms. I let you do the details.