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Question Number 63190 by Tawa1 last updated on 30/Jun/19

Test its convergence: Σ_(n = 1) ^∞ (1/(n^3 sin^2 n))

$$\mathrm{Test}\:\mathrm{its}\:\mathrm{convergence}:\:\:\:\:\:\:\:\:\underset{\mathrm{n}\:=\:\mathrm{1}} {\overset{\infty} {\sum}}\:\frac{\mathrm{1}}{\mathrm{n}^{\mathrm{3}} \:\mathrm{sin}^{\mathrm{2}} \mathrm{n}} \\ $$

Commented by mathmax by abdo last updated on 01/Jul/19

let S = Σ_(n=1) ^∞ (1/(n^3 sin^2 n)) ⇒ S =Σ_(n=1) ^∞ (1/(n^3 (((1−cos(2n))/2)))) = 2 Σ_(n=1) ^∞ (1/(n^3 (1−cos(2n))) we have ∣ ∣a∣−∣b∣∣ ≤∣a−b∣ ≤∣a∣ +∣b∣ ⇒ ∣1−∣cos(2n)∣≤∣1−cos(2n)∣≤1+∣cos(2n)∣≤2 ⇒ (1/2) ≤ (1/(∣1−cos(2n)∣))≤ (1/(∣1−∣cos(2n)∣∣)) ⇒ (1/n^3 ) ≤ (1/(n^3 ∣1−cos(2n)∣)) ≤(1/(n^3 ∣1−∣cos(2n)∣∣)) ⇒ Σ_(n=1) ^∞ (2/n^3 ) ≤ S ≤ Σ_(n=1) ^∞ (2/(n^3 ∣1−∣cos(2n)∣)) the sequence n→(2/(∣1−∣cos(2n)∣∣)) is borned ⇒ ∃ M_n >0 .∃n_o integr / n>n_0 ⇒ (2/(∣1−∣cos(2n)∣)) < M_n ⇒ Σ_(n=1) ^∞ (2/n^3 ) ≤ S ≤Σ_(n=1) ^∞ (M_n /n^3 ) the 2 series are convergents ⇒ S conerges .

$${let}\:{S}\:=\:\sum_{{n}=\mathrm{1}} ^{\infty} \:\frac{\mathrm{1}}{{n}^{\mathrm{3}} \:{sin}^{\mathrm{2}} {n}}\:\Rightarrow\:{S}\:=\sum_{{n}=\mathrm{1}} ^{\infty} \:\frac{\mathrm{1}}{{n}^{\mathrm{3}} \left(\frac{\mathrm{1}−{cos}\left(\mathrm{2}{n}\right)}{\mathrm{2}}\right)} \\ $$$$=\:\mathrm{2}\:\sum_{{n}=\mathrm{1}} ^{\infty} \:\:\frac{\mathrm{1}}{{n}^{\mathrm{3}} \left(\mathrm{1}−{cos}\left(\mathrm{2}{n}\right)\right.} \\ $$$$\:{we}\:{have}\:\:\mid\:\mid{a}\mid−\mid{b}\mid\mid\:\leqslant\mid{a}−{b}\mid\:\leqslant\mid{a}\mid\:+\mid{b}\mid\:\Rightarrow\:\mid\mathrm{1}−\mid{cos}\left(\mathrm{2}{n}\right)\mid\leqslant\mid\mathrm{1}−{cos}\left(\mathrm{2}{n}\right)\mid\leqslant\mathrm{1}+\mid{cos}\left(\mathrm{2}{n}\right)\mid\leqslant\mathrm{2}\:\Rightarrow \\ $$$$\frac{\mathrm{1}}{\mathrm{2}}\:\leqslant\:\frac{\mathrm{1}}{\mid\mathrm{1}−{cos}\left(\mathrm{2}{n}\right)\mid}\leqslant\:\frac{\mathrm{1}}{\mid\mathrm{1}−\mid{cos}\left(\mathrm{2}{n}\right)\mid\mid}\:\Rightarrow \\ $$$$\frac{\mathrm{1}}{{n}^{\mathrm{3}} }\:\leqslant\:\frac{\mathrm{1}}{{n}^{\mathrm{3}} \mid\mathrm{1}−{cos}\left(\mathrm{2}{n}\right)\mid}\:\leqslant\frac{\mathrm{1}}{{n}^{\mathrm{3}} \mid\mathrm{1}−\mid{cos}\left(\mathrm{2}{n}\right)\mid\mid}\:\Rightarrow\: \\ $$$$\sum_{{n}=\mathrm{1}} ^{\infty} \:\:\frac{\mathrm{2}}{{n}^{\mathrm{3}} }\:\leqslant\:\:{S}\:\leqslant\:\:\sum_{{n}=\mathrm{1}} ^{\infty} \:\:\:\frac{\mathrm{2}}{{n}^{\mathrm{3}} \mid\mathrm{1}−\mid{cos}\left(\mathrm{2}{n}\right)\mid} \\ $$$${the}\:{sequence}\:\:{n}\rightarrow\frac{\mathrm{2}}{\mid\mathrm{1}−\mid{cos}\left(\mathrm{2}{n}\right)\mid\mid}\:{is}\:{borned}\:\Rightarrow\:\exists\:{M}_{{n}} \:>\mathrm{0}\:\:.\exists{n}_{{o}} \:{integr}\:/ \\ $$$${n}>{n}_{\mathrm{0}} \Rightarrow\:\frac{\mathrm{2}}{\mid\mathrm{1}−\mid{cos}\left(\mathrm{2}{n}\right)\mid}\:<\:\:{M}_{{n}} \:\Rightarrow\:\sum_{{n}=\mathrm{1}} ^{\infty} \:\frac{\mathrm{2}}{{n}^{\mathrm{3}} }\:\leqslant\:{S}\:\leqslant\sum_{{n}=\mathrm{1}} ^{\infty} \:\:\:\frac{{M}_{{n}} }{{n}^{\mathrm{3}} } \\ $$$${the}\:\mathrm{2}\:{series}\:{are}\:{convergents}\:\Rightarrow\:{S}\:\:{conerges}\:. \\ $$

Commented by Tawa1 last updated on 01/Jul/19

God bless you sir

$$\mathrm{God}\:\mathrm{bless}\:\mathrm{you}\:\mathrm{sir} \\ $$

Commented by mathmax by abdo last updated on 01/Jul/19

you are welcome.

$${you}\:{are}\:{welcome}. \\ $$

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