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Question Number 106246 by mohammad17 last updated on 03/Aug/20

Commented by mohammad17 last updated on 03/Aug/20

test the series converge or diverge

$${test}\:{the}\:{series}\:{converge}\:{or}\:{diverge} \\ $$

Commented by mathmax by abdo last updated on 04/Aug/20

9) S =Σ_(n=1) ^∞ (−1)^n u_n with u_n =(((n+5)^2 )/((n+3)^5 )) =ϕ(n) with ϕ(x) =(((x+5)^2 )/((x+3)^5 )) ( we take x>0) ⇒ ϕ^′ (x) =((2(x+5)(x+3)^5 −5(x+3)^4 (x+5)^2 )/((x+3)^(10) )) =((2(x+5)(x+3)−5(x+5)^2 )/((x+3)^6 )) =((2x^2 +6x +10x+30−5(x^2 +10x+25))/((x+3)^6 )) =((2x^2 +16x +30−5x^2 −50x −125)/((x+3)^6 )) =((−3x^2 −34x−125)/((x+3)^2 ))<0 ⇒ϕ is decreazing ⇒S is alternate serie convergent .

$$\left.\mathrm{9}\right)\:\mathrm{S}\:=\sum_{\mathrm{n}=\mathrm{1}} ^{\infty} \:\left(−\mathrm{1}\right)^{\mathrm{n}} \:\mathrm{u}_{\mathrm{n}} \:\mathrm{with}\:\mathrm{u}_{\mathrm{n}} =\frac{\left(\mathrm{n}+\mathrm{5}\right)^{\mathrm{2}} }{\left(\mathrm{n}+\mathrm{3}\right)^{\mathrm{5}} }\:=\varphi\left(\mathrm{n}\right)\:\mathrm{with} \\ $$$$\varphi\left(\mathrm{x}\right)\:=\frac{\left(\mathrm{x}+\mathrm{5}\right)^{\mathrm{2}} }{\left(\mathrm{x}+\mathrm{3}\right)^{\mathrm{5}} }\:\left(\:\mathrm{we}\:\mathrm{take}\:\mathrm{x}>\mathrm{0}\right)\:\Rightarrow \\ $$$$\varphi^{'} \left(\mathrm{x}\right)\:=\frac{\mathrm{2}\left(\mathrm{x}+\mathrm{5}\right)\left(\mathrm{x}+\mathrm{3}\right)^{\mathrm{5}} −\mathrm{5}\left(\mathrm{x}+\mathrm{3}\right)^{\mathrm{4}} \left(\mathrm{x}+\mathrm{5}\right)^{\mathrm{2}} }{\left(\mathrm{x}+\mathrm{3}\right)^{\mathrm{10}} } \\ $$$$=\frac{\mathrm{2}\left(\mathrm{x}+\mathrm{5}\right)\left(\mathrm{x}+\mathrm{3}\right)−\mathrm{5}\left(\mathrm{x}+\mathrm{5}\right)^{\mathrm{2}} }{\left(\mathrm{x}+\mathrm{3}\right)^{\mathrm{6}} }\:=\frac{\mathrm{2x}^{\mathrm{2}} +\mathrm{6x}\:+\mathrm{10x}+\mathrm{30}−\mathrm{5}\left(\mathrm{x}^{\mathrm{2}} +\mathrm{10x}+\mathrm{25}\right)}{\left(\mathrm{x}+\mathrm{3}\right)^{\mathrm{6}} } \\ $$$$=\frac{\mathrm{2x}^{\mathrm{2}} \:+\mathrm{16x}\:+\mathrm{30}−\mathrm{5x}^{\mathrm{2}} −\mathrm{50x}\:−\mathrm{125}}{\left(\mathrm{x}+\mathrm{3}\right)^{\mathrm{6}} } \\ $$$$=\frac{−\mathrm{3x}^{\mathrm{2}} −\mathrm{34x}−\mathrm{125}}{\left(\mathrm{x}+\mathrm{3}\right)^{\mathrm{2}} }<\mathrm{0}\:\Rightarrow\varphi\:\mathrm{is}\:\mathrm{decreazing}\:\Rightarrow\mathrm{S}\:\mathrm{is}\:\mathrm{alternate}\:\mathrm{serie} \\ $$$$\mathrm{convergent}\:\:. \\ $$$$ \\ $$

Commented by mohammad17 last updated on 04/Aug/20

thank you sir

$${thank}\:{you}\:{sir} \\ $$

Answered by mathmax by abdo last updated on 03/Aug/20

3) S =Σ_(n=1) ^∞ (((−1)^n )/(n+2^n )) for all n≥1 n+2^n >2^(n ) ⇒(1/(n+2^n ))<(1/2^n ) ⇒ ∣(((−1)^n )/(n+2^n ))∣<(1/2^n ) and the serie Σ (1/2^n ) converges ⇒ S converves

$$\left.\mathrm{3}\right)\:\mathrm{S}\:=\sum_{\mathrm{n}=\mathrm{1}} ^{\infty} \:\frac{\left(−\mathrm{1}\right)^{\mathrm{n}} }{\mathrm{n}+\mathrm{2}^{\mathrm{n}} }\:\:\:\mathrm{for}\:\mathrm{all}\:\mathrm{n}\geqslant\mathrm{1}\:\:\:\mathrm{n}+\mathrm{2}^{\mathrm{n}} \:>\mathrm{2}^{\mathrm{n}\:} \:\Rightarrow\frac{\mathrm{1}}{\mathrm{n}+\mathrm{2}^{\mathrm{n}} }<\frac{\mathrm{1}}{\mathrm{2}^{\mathrm{n}} }\:\Rightarrow \\ $$$$\mid\frac{\left(−\mathrm{1}\right)^{\mathrm{n}} }{\mathrm{n}+\mathrm{2}^{\mathrm{n}} }\mid<\frac{\mathrm{1}}{\mathrm{2}^{\mathrm{n}} }\:\:\mathrm{and}\:\mathrm{the}\:\mathrm{serie}\:\Sigma\:\frac{\mathrm{1}}{\mathrm{2}^{\mathrm{n}} }\:\mathrm{converges}\:\Rightarrow\:\mathrm{S}\:\mathrm{converves} \\ $$

Commented by mohammad17 last updated on 04/Aug/20

thank you sir

$${thank}\:{you}\:{sir} \\ $$

Commented by mathmax by abdo last updated on 04/Aug/20

you are welcome sir

$$\mathrm{you}\:\mathrm{are}\:\mathrm{welcome}\:\mathrm{sir} \\ $$

Answered by mathmax by abdo last updated on 03/Aug/20

5) S =Σ_(n=2) ^∞ (((−1)^(n−1) )/((nln(n))^2 )) ⇒ S =Σ_(n=2) ^∞ (−1)^(n−1) u_n u_n =(1/((nln(n))^2 )) u_n is >0 decrezing to 0 ⇒ S is a alternate serie convergente another way ∣(((−1)^(n−1) )/((nlnn)^2 ))∣≤(1/((nlnn)^2 )) the sdquence u_n =(1/((nlnn)^2 )) is decreazing ⇒Σ_(n=2) ^∞ (1/((nlnn)^2 )) and ∫_2 ^(+∞) (dx/((xlnx)^2 )) have same nature of convergence we have ∫_2 ^(+∞) (dx/((xlnx)^2 )) =_(lnx=t) ∫_(ln2) ^(+∞) ((e^t dt)/((e^t t)^2 )) =∫_(ln2) ^(+∞) (e^(−t) /t^2 ) dt and this integral is convergent ⇒ S is convergent

$$\left.\mathrm{5}\right)\:\mathrm{S}\:=\sum_{\mathrm{n}=\mathrm{2}} ^{\infty} \:\frac{\left(−\mathrm{1}\right)^{\mathrm{n}−\mathrm{1}} }{\left(\mathrm{nln}\left(\mathrm{n}\right)\right)^{\mathrm{2}} }\:\Rightarrow\:\mathrm{S}\:=\sum_{\mathrm{n}=\mathrm{2}} ^{\infty} \:\left(−\mathrm{1}\right)^{\mathrm{n}−\mathrm{1}} \:\mathrm{u}_{\mathrm{n}} \\ $$$$\mathrm{u}_{\mathrm{n}} =\frac{\mathrm{1}}{\left(\mathrm{nln}\left(\mathrm{n}\right)\right)^{\mathrm{2}} }\:\:\mathrm{u}_{\mathrm{n}} \mathrm{is}\:>\mathrm{0}\:\:\:\mathrm{decrezing}\:\mathrm{to}\:\mathrm{0}\:\Rightarrow\:\mathrm{S}\:\mathrm{is}\:\mathrm{a}\:\mathrm{alternate}\:\mathrm{serie} \\ $$$$\mathrm{convergente} \\ $$$$\mathrm{another}\:\mathrm{way}\:\:\mid\frac{\left(−\mathrm{1}\right)^{\mathrm{n}−\mathrm{1}} }{\left(\mathrm{nlnn}\right)^{\mathrm{2}} }\mid\leqslant\frac{\mathrm{1}}{\left(\mathrm{nlnn}\right)^{\mathrm{2}} }\:\:\mathrm{the}\:\mathrm{sdquence}\:\mathrm{u}_{\mathrm{n}} =\frac{\mathrm{1}}{\left(\mathrm{nlnn}\right)^{\mathrm{2}} } \\ $$$$\mathrm{is}\:\mathrm{decreazing}\:\Rightarrow\sum_{\mathrm{n}=\mathrm{2}} ^{\infty} \:\frac{\mathrm{1}}{\left(\mathrm{nlnn}\right)^{\mathrm{2}} }\:\mathrm{and}\:\int_{\mathrm{2}} ^{+\infty} \:\frac{\mathrm{dx}}{\left(\mathrm{xlnx}\right)^{\mathrm{2}} }\:\mathrm{have}\:\mathrm{same} \\ $$$$\mathrm{nature}\:\mathrm{of}\:\mathrm{convergence}\:\:\mathrm{we}\:\mathrm{have} \\ $$$$\int_{\mathrm{2}} ^{+\infty} \:\frac{\mathrm{dx}}{\left(\mathrm{xlnx}\right)^{\mathrm{2}} }\:=_{\mathrm{lnx}=\mathrm{t}} \:\:\:\:\int_{\mathrm{ln2}} ^{+\infty} \:\frac{\mathrm{e}^{\mathrm{t}} \mathrm{dt}}{\left(\mathrm{e}^{\mathrm{t}} \mathrm{t}\right)^{\mathrm{2}} }\:=\int_{\mathrm{ln2}} ^{+\infty} \:\frac{\mathrm{e}^{−\mathrm{t}} }{\mathrm{t}^{\mathrm{2}} }\:\mathrm{dt}\:\:\mathrm{and}\:\mathrm{this}\:\mathrm{integral}\:\mathrm{is} \\ $$$$\mathrm{convergent}\:\Rightarrow\:\mathrm{S}\:\mathrm{is}\:\mathrm{convergent} \\ $$

Commented by mohammad17 last updated on 04/Aug/20

thank you sir

$${thank}\:{you}\:{sir} \\ $$

Commented by mathmax by abdo last updated on 04/Aug/20

you are welcome sir

$$\mathrm{you}\:\mathrm{are}\:\mathrm{welcome}\:\mathrm{sir} \\ $$

Answered by mathmax by abdo last updated on 04/Aug/20

7) S =Σ_(n=1) ^(∞ ) ((sin(αn))/(n!)) ⇒ S =Σ_(n=1) ^∞ u_n ∣(u_(n+1) /u_n )∣ =∣((sin(α(n+1)))/((n+1)!))×((n!)/(sin(αn)))∣ =∣((sin(αn+n))/(sin(αn)))∣×(1/(n+1)) →0(n→+∞) S is convervent another way we have ∣((sin(αn))/(n!))∣≤(1/(n!)) ∀n and Σ(1/(n!)) converges ⇒ S converges absolument ⇒S conerges..

$$\left.\mathrm{7}\right)\:\mathrm{S}\:=\sum_{\mathrm{n}=\mathrm{1}} ^{\infty\:\:} \:\frac{\mathrm{sin}\left(\alpha\mathrm{n}\right)}{\mathrm{n}!}\:\Rightarrow\:\mathrm{S}\:=\sum_{\mathrm{n}=\mathrm{1}} ^{\infty} \:\mathrm{u}_{\mathrm{n}} \\ $$$$\mid\frac{\mathrm{u}_{\mathrm{n}+\mathrm{1}} }{\mathrm{u}_{\mathrm{n}} }\mid\:=\mid\frac{\mathrm{sin}\left(\alpha\left(\mathrm{n}+\mathrm{1}\right)\right)}{\left(\mathrm{n}+\mathrm{1}\right)!}×\frac{\mathrm{n}!}{\mathrm{sin}\left(\alpha\mathrm{n}\right)}\mid\:=\mid\frac{\mathrm{sin}\left(\alpha\mathrm{n}+\mathrm{n}\right)}{\mathrm{sin}\left(\alpha\mathrm{n}\right)}\mid×\frac{\mathrm{1}}{\mathrm{n}+\mathrm{1}}\:\rightarrow\mathrm{0}\left(\mathrm{n}\rightarrow+\infty\right) \\ $$$$\mathrm{S}\:\mathrm{is}\:\mathrm{convervent}\: \\ $$$$\mathrm{another}\:\mathrm{way}\:\:\mathrm{we}\:\mathrm{have}\:\mid\frac{\mathrm{sin}\left(\alpha\mathrm{n}\right)}{\mathrm{n}!}\mid\leqslant\frac{\mathrm{1}}{\mathrm{n}!}\:\:\forall\mathrm{n}\:\:\mathrm{and}\:\Sigma\frac{\mathrm{1}}{\mathrm{n}!}\:\mathrm{converges}\:\Rightarrow \\ $$$$\mathrm{S}\:\mathrm{converges}\:\mathrm{absolument}\:\Rightarrow\mathrm{S}\:\mathrm{conerges}.. \\ $$

Commented by mohammad17 last updated on 04/Aug/20

thank you sir

$${thank}\:{you}\:{sir}\: \\ $$

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