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Question Number 38898 by math khazana by abdo last updated on 01/Jul/18

find nature od the seri Σ_(n=1) ^∞   ((n(−1)^([x]) )/(1+n[x]^3 ))

$${find}\:{nature}\:{od}\:{the}\:{seri}\:\sum_{{n}=\mathrm{1}} ^{\infty} \:\:\frac{{n}\left(−\mathrm{1}\right)^{\left[{x}\right]} }{\mathrm{1}+{n}\left[{x}\right]^{\mathrm{3}} } \\ $$

Commented by math khazana by abdo last updated on 01/Jul/18

S =lim_(n→+∞)  S_n  with S_n = Σ_(k=1) ^(n−1)   ∫_k ^(k+1)   ((n(−1)^k )/(1+nk^3 ))dx  =Σ_(k=1) ^(n−1)  n(−1)^k  (1/(1+nk^3 )) ⇒∣S_n ∣ ≤Σ_(k=1) ^(n−1)  (n/(1+nk^3 ))  ≤Σ_(k=1) ^n   (1/k^3 ) =ξ_(n−1) (3) ⇒ ∣S∣≤ ξ(3) ⇒ S is convergent.

$${S}\:={lim}_{{n}\rightarrow+\infty} \:{S}_{{n}} \:{with}\:{S}_{{n}} =\:\sum_{{k}=\mathrm{1}} ^{{n}−\mathrm{1}} \:\:\int_{{k}} ^{{k}+\mathrm{1}} \:\:\frac{{n}\left(−\mathrm{1}\right)^{{k}} }{\mathrm{1}+{nk}^{\mathrm{3}} }{dx} \\ $$$$=\sum_{{k}=\mathrm{1}} ^{{n}−\mathrm{1}} \:{n}\left(−\mathrm{1}\right)^{{k}} \:\frac{\mathrm{1}}{\mathrm{1}+{nk}^{\mathrm{3}} }\:\Rightarrow\mid{S}_{{n}} \mid\:\leqslant\sum_{{k}=\mathrm{1}} ^{{n}−\mathrm{1}} \:\frac{{n}}{\mathrm{1}+{nk}^{\mathrm{3}} } \\ $$$$\leqslant\sum_{{k}=\mathrm{1}} ^{{n}} \:\:\frac{\mathrm{1}}{{k}^{\mathrm{3}} }\:=\xi_{{n}−\mathrm{1}} \left(\mathrm{3}\right)\:\Rightarrow\:\mid{S}\mid\leqslant\:\xi\left(\mathrm{3}\right)\:\Rightarrow\:{S}\:{is}\:{convergent}. \\ $$

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