Question

For problems 5 and 6 find the radius of convergence and the interval of convergence 5. 5 (x-3)n V n=0 n3n TS r3n+1 (2n - 1)(2n +1)

Answer 1

1. 
   į Zan (let) A5) .hdi = 3an (en 120 - nan test. Ratio 1 132 ] e ihn , anti T(2-3)2t) Tintillanti) _31 nep I do (2-3) noop (n+1)3 - litto ze ☆ = - Cho to be hard |</ >
   
   Itse R² 3 ] Radius of convergence -3x | x-31 <+3 0<x< 6 {divide by 3 } { adding 3 atfe=o q (-3) + x = 6 nzon noon Bnc h Harmonic series are divergent so, n sa пер Usted Alternating Serme test lim alanda lim 16 = 0 Convergence Intervallo = x <6) or [0,6)
   
   3ntl 6) to co-1) en +1) = Ean (let) Using Ratio test 318+1)+] (2n-1) (2n+1) 233 + . anti lalim ? I no an 1126171)-1)(260 +1 +1 R. reint . 3ntl R .to to top =*by h 94) 2t n= 0 So, berkel [R=1] radius of convergence 1x1 <1 -1 <r<l -> -1<x</ art & 2n iscentis | at(n = +1 (-03n+/ heto (2n-1) (n +1) Using Diverges test 1 Using aternating Sesue tert lim analim dos c2n-1) (2n+1) | din o anl = limon can-1) (2n +1) d o Camergas) = = = 0 (comierous). Interual le nel or El, l ] Radius [R=1 insa