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Question-145450

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Question Number 145450 by solihin last updated on 05/Jul/21
Answered by Olaf_Thorendsen last updated on 05/Jul/21
a) a_n = (2^n /3^(n+1) ), n∈N^∗ a_n =(1/3)((2/3))^n →_∞ 0 b) S = Σ_(n=1) ^∞ (((−1)^n n^2 )/(3n^4 +2n^3 +5n−2)) S = Σ(−1)^n a_n with ∣a_n ∣→_∞ 0 and ∣(a_(n+1) /a_n )∣ < 1 ⇒ convergence and Σ∣a_n ∣ = Σ_(n=1) ^∞ (n^2 /(3n^4 +2n^3 +5n−2)) ≤ Σ_(n=1) ^∞ (n^2 /(3n^4 )) = (1/3)Σ_(n=1) ^∞ (1/n^2 ) = (π^2 /(18)) S is absolutely convergent
$$\left.{a}\right) \\ $$$${a}_{{n}} \:=\:\frac{\mathrm{2}^{{n}} }{\mathrm{3}^{{n}+\mathrm{1}} },\:{n}\in\mathbb{N}^{\ast} \\ $$$${a}_{{n}} \:=\frac{\mathrm{1}}{\mathrm{3}}\left(\frac{\mathrm{2}}{\mathrm{3}}\right)^{{n}} \underset{\infty} {\rightarrow}\mathrm{0} \\ $$$$\left.{b}\right) \\ $$$$\mathrm{S}\:=\:\underset{{n}=\mathrm{1}} {\overset{\infty} {\sum}}\frac{\left(−\mathrm{1}\right)^{{n}} {n}^{\mathrm{2}} }{\mathrm{3}{n}^{\mathrm{4}} +\mathrm{2}{n}^{\mathrm{3}} +\mathrm{5}{n}−\mathrm{2}} \\ $$$$\mathrm{S}\:=\:\Sigma\left(−\mathrm{1}\right)^{{n}} {a}_{{n}} \\ $$$${with}\:\mid{a}_{{n}} \mid\underset{\infty} {\rightarrow}\mathrm{0} \\ $$$$\mathrm{and}\:\mid\frac{{a}_{{n}+\mathrm{1}} }{{a}_{{n}} }\mid\:<\:\mathrm{1} \\ $$$$\Rightarrow\:\mathrm{convergence} \\ $$$$\mathrm{and}\:\Sigma\mid{a}_{{n}} \mid\:=\:\underset{{n}=\mathrm{1}} {\overset{\infty} {\sum}}\frac{{n}^{\mathrm{2}} }{\mathrm{3}{n}^{\mathrm{4}} +\mathrm{2}{n}^{\mathrm{3}} +\mathrm{5}{n}−\mathrm{2}} \\ $$$$\leqslant\:\underset{{n}=\mathrm{1}} {\overset{\infty} {\sum}}\frac{{n}^{\mathrm{2}} }{\mathrm{3}{n}^{\mathrm{4}} }\:=\:\frac{\mathrm{1}}{\mathrm{3}}\underset{{n}=\mathrm{1}} {\overset{\infty} {\sum}}\frac{\mathrm{1}}{{n}^{\mathrm{2}} }\:=\:\frac{\pi^{\mathrm{2}} }{\mathrm{18}} \\ $$$$\mathrm{S}\:\mathrm{is}\:\mathrm{absolutely}\:\mathrm{convergent} \\ $$
Answered by mathmax by abdo last updated on 05/Jul/21
a_n =(2^n /3^(n+1) ) ⇒a_n =(1/3)((2/3))^n we have ∣(2/3)∣<1 ⇒lim_(n→+∞) a_n =0 b)Σ u_n with u_n =(((−1)^n n^2 )/(3n^4 +2n^3 +5n−2)) ⇒∣u_n ∣∼(n^2 /(3n^4 ))=(1/(3n^2 )) ⇒Σ u_n converge absolument c)Σ v_n with v_n =(5n)!(x−2)^(n ) ⇒ (^n (√(∣v_n ∣)))=((5n)!)^(1/n) ∣x−2∣ ((5n)!)^(1/n) =e^((1/n)log(5n)!) we have n!∼n^n e^(−n) (√(2πn)) ⇒ )5n)!∼(5n)^(5n) e^(−5n) (√(10πn)) ⇒ log(5n!)∼5nlog(5n)−5n+(1/2)log(10πn) ⇒ ((log(5n!))/n)∼5log(5n)−5+(1/(2n))log(10πn)→+∞ so for x≠2 (∣v_n ∣)^(1/n) →∞ ⇒Σ v_n diverges...!
$$\mathrm{a}_{\mathrm{n}} =\frac{\mathrm{2}^{\mathrm{n}} }{\mathrm{3}^{\mathrm{n}+\mathrm{1}} }\:\Rightarrow\mathrm{a}_{\mathrm{n}} =\frac{\mathrm{1}}{\mathrm{3}}\left(\frac{\mathrm{2}}{\mathrm{3}}\right)^{\mathrm{n}} \:\:\:\mathrm{we}\:\mathrm{have}\:\mid\frac{\mathrm{2}}{\mathrm{3}}\mid<\mathrm{1}\:\Rightarrow\mathrm{lim}_{\mathrm{n}\rightarrow+\infty} \mathrm{a}_{\mathrm{n}} =\mathrm{0} \\ $$$$\left.\mathrm{b}\right)\Sigma\:\mathrm{u}_{\mathrm{n}} \:\:\mathrm{with}\:\mathrm{u}_{\mathrm{n}} =\frac{\left(−\mathrm{1}\right)^{\mathrm{n}} \mathrm{n}^{\mathrm{2}} }{\mathrm{3n}^{\mathrm{4}} \:+\mathrm{2n}^{\mathrm{3}} \:+\mathrm{5n}−\mathrm{2}}\:\Rightarrow\mid\mathrm{u}_{\mathrm{n}} \mid\sim\frac{\mathrm{n}^{\mathrm{2}} }{\mathrm{3n}^{\mathrm{4}} }=\frac{\mathrm{1}}{\mathrm{3n}^{\mathrm{2}} } \\ $$$$\Rightarrow\Sigma\:\mathrm{u}_{\mathrm{n}} \mathrm{converge}\:\mathrm{absolument} \\ $$$$\left.\mathrm{c}\right)\Sigma\:\mathrm{v}_{\mathrm{n}} \:\:\mathrm{with}\:\mathrm{v}_{\mathrm{n}} =\left(\mathrm{5n}\right)!\left(\mathrm{x}−\mathrm{2}\right)^{\mathrm{n}\:} \:\Rightarrow \\ $$$$\left(^{\mathrm{n}} \sqrt{\mid\mathrm{v}_{\mathrm{n}} \mid}\right)=\left(\left(\mathrm{5n}\right)!\right)^{\frac{\mathrm{1}}{\mathrm{n}}} \mid\mathrm{x}−\mathrm{2}\mid \\ $$$$\left(\left(\mathrm{5n}\right)!\right)^{\frac{\mathrm{1}}{\mathrm{n}}} \:=\mathrm{e}^{\frac{\mathrm{1}}{\mathrm{n}}\mathrm{log}\left(\mathrm{5n}\right)!} \:\mathrm{we}\:\mathrm{have}\:\mathrm{n}!\sim\mathrm{n}^{\mathrm{n}} \:\mathrm{e}^{−\mathrm{n}} \sqrt{\mathrm{2}\pi\mathrm{n}}\:\Rightarrow \\ $$$$\left.\right)\left.\mathrm{5n}\right)!\sim\left(\mathrm{5n}\right)^{\mathrm{5n}} \:\mathrm{e}^{−\mathrm{5n}} \sqrt{\mathrm{10}\pi\mathrm{n}}\:\Rightarrow \\ $$$$\mathrm{log}\left(\mathrm{5n}!\right)\sim\mathrm{5nlog}\left(\mathrm{5n}\right)−\mathrm{5n}+\frac{\mathrm{1}}{\mathrm{2}}\mathrm{log}\left(\mathrm{10}\pi\mathrm{n}\right)\:\Rightarrow \\ $$$$\frac{\mathrm{log}\left(\mathrm{5n}!\right)}{\mathrm{n}}\sim\mathrm{5log}\left(\mathrm{5n}\right)−\mathrm{5}+\frac{\mathrm{1}}{\mathrm{2n}}\mathrm{log}\left(\mathrm{10}\pi\mathrm{n}\right)\rightarrow+\infty\:\mathrm{so}\:\mathrm{for}\:\mathrm{x}\neq\mathrm{2} \\ $$$$\left(\mid\mathrm{v}_{\mathrm{n}} \mid\right)^{\frac{\mathrm{1}}{\mathrm{n}}} \:\rightarrow\infty\:\Rightarrow\Sigma\:\mathrm{v}_{\mathrm{n}} \:\mathrm{diverges}…! \\ $$

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