Question Number 113897 by Lordose last updated on 16/Sep/20
Answered by mathmax by abdo last updated on 17/Sep/20
![let u_n =(3^n /( (√((3n−2)2^n )))) ⇒u_n =((((3/( (√2))))^n )/( (√(3n−2)))) (n≥1) ⇒ (u_(n+1) /u_n ) =((((3/( (√2))))^(n+1) )/( (√(3n+1))))×((√(3n−2))/(((3/( (√2))))^n )) =(√((3n−2)/(3n+1)))×((3/( (√2)))) and lim_(n→+∞) (u_(n+1) /u_n ) =(3/( (√2))) so for ∣x∣ <(3/( (√2))) the serie converges and for ∣x∣≥(3/( (√2))) the serie diverges finally I_c =]−(3/( (√2))),(3/( (√2)))[](https://www.tinkutara.com/question/Q114059.png)
$$\mathrm{let}\:\mathrm{u}_{\mathrm{n}} =\frac{\mathrm{3}^{\mathrm{n}} }{\:\sqrt{\left(\mathrm{3n}−\mathrm{2}\right)\mathrm{2}^{\mathrm{n}} }}\:\Rightarrow\mathrm{u}_{\mathrm{n}} =\frac{\left(\frac{\mathrm{3}}{\:\sqrt{\mathrm{2}}}\right)^{\mathrm{n}} }{\:\sqrt{\mathrm{3n}−\mathrm{2}}}\:\:\left(\mathrm{n}\geqslant\mathrm{1}\right)\:\Rightarrow \\ $$$$\frac{\mathrm{u}_{\mathrm{n}+\mathrm{1}} }{\mathrm{u}_{\mathrm{n}} }\:=\frac{\left(\frac{\mathrm{3}}{\:\sqrt{\mathrm{2}}}\right)^{\mathrm{n}+\mathrm{1}} }{\:\sqrt{\mathrm{3n}+\mathrm{1}}}×\frac{\sqrt{\mathrm{3n}−\mathrm{2}}}{\left(\frac{\mathrm{3}}{\:\sqrt{\mathrm{2}}}\right)^{\mathrm{n}} }\:=\sqrt{\frac{\mathrm{3n}−\mathrm{2}}{\mathrm{3n}+\mathrm{1}}}×\left(\frac{\mathrm{3}}{\:\sqrt{\mathrm{2}}}\right)\:\:\mathrm{and}\:\mathrm{lim}_{\mathrm{n}\rightarrow+\infty} \:\frac{\mathrm{u}_{\mathrm{n}+\mathrm{1}} }{\mathrm{u}_{\mathrm{n}} } \\ $$$$=\frac{\mathrm{3}}{\:\sqrt{\mathrm{2}}}\:\:\mathrm{so}\:\mathrm{for}\:\mid\mathrm{x}\mid\:<\frac{\mathrm{3}}{\:\sqrt{\mathrm{2}}}\:\:\mathrm{the}\:\mathrm{serie}\:\mathrm{converges}\:\mathrm{and}\:\mathrm{for}\:\mid\mathrm{x}\mid\geqslant\frac{\mathrm{3}}{\:\sqrt{\mathrm{2}}} \\ $$$$\left.\mathrm{the}\:\mathrm{serie}\:\mathrm{diverges}\:\mathrm{finally}\:\mathrm{I}_{\mathrm{c}} =\right]−\frac{\mathrm{3}}{\:\sqrt{\mathrm{2}}},\frac{\mathrm{3}}{\:\sqrt{\mathrm{2}}}\left[\right. \\ $$