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Question Number 119212 by help last updated on 23/Oct/20

Answered by Olaf last updated on 23/Oct/20

∀k∈N^∗ , (k−1)^3 −(k^3 +4) = −3k^2 +3k−5 < 0 ⇒ (k−1)^3 > k^3 +4 and : k≥2, ((k^2 −1)/(k^3 +4)) ≥ ((k^2 −1)/((k−1)^3 )) = ((k+1)/((k−1)^2 )) ≥ ((k+1)/((k+1)^2 )) = (1/(k+1)) ⇒ Σ_(k=2) ^∞ ((k^2 −1)/(k^3 +4)) ≥ Σ_(k=2) ^∞ (1/(k+1)) = Σ_(k=3) ^∞ (1/k) The harmonic serie Σ_(k=1) ^n (1/k) is divergent. ⇒ Σ_(k=1) ^∞ ((k^2 −1)/(k^3 +4)) is divergent.

∀k∈N∗,(k−1)3−(k3+4)=−3k2+3k−5<0⇒(k−1)3>k3+4and:k⩾2,k2−1k3+4⩾k2−1(k−1)3=k+1(k−1)2⩾k+1(k+1)2=1k+1⇒∑∞k=2k2−1k3+4⩾∑∞k=21k+1=∑∞k=31kTheharmonicserie∑nk=11kisdivergent.⇒∑∞k=1k2−1k3+4isdivergent.

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