Question

number 4 as clearly as possible

1. Find the limit of the following sequences (find lim an) n n a.) an = n +3 b.) an = V35n n- 2. Determine whether the following series converge or diverge. -3 (n + 2)n + 5 b.) tan-'(n) n2 + 1 a.) 5 nel 3. Determine the radius of convergence and the interval of convergence of the series 2" (x – 3)" n n=1 n=0 (-1)", 2n 4. Using the power series cos(x) (2n)! (-« < x < 0o), to find a power series for the function f(x) = sin(x) sin(3.c) and its interval of convergence. 5. Find the power series representation for the function f(x) = and its interval (3.x - 2)2 of convergence. 6. Find the power series representation for the function f(x) = (1 - 2x)3 and its interval of convergence. 2.c 7. Find a power series for the function f(x) = 3x2 - 52 - 2 and its interval of convergence. 8. Given f(x) = ln(1 – 3x), without using any well known power series, find the MacLaurin series of f(x) and its interval of convergence.

Answer 1

nzo Now V O 6 2n (-1)^ (4x) 21 2 [ (2n)! n=o n=0 20-240 4) we we have cos(x) = Ž (-1) 202 (-0<x<a Too) Sin (2) Sin (3x) = [ cos(2x) – cos (4x)] f(x) = sin(n) (sin 3x) = { [eoskan)-2054(0)] 3 (-1)^ (2x) I (2n): E (1)? x 2] (2n)? (-1)? (222 27").a. 212) (-)" (22190) 212n)! 2(n+) 4 (n+1) (an)! lim (2n+2)! an an an 2.a اح n=0 (21)! n= Z M SO, an= 2 anti = lim X R= an n72 no lim x 2n 2 229n 2.2.4 - 2016 (2n! (2n+2)(2n+1Dan! AC gan_24h) + 12.24 (229241) (2n+2)(2011) an lim n70 A lin 12.249 u nD 2n+2)(20+) * 229 241) (2n+2) (2011)

as the denominator is bigger than numerator, as n> denominator goes to G so the lim 12.24n 1 precisely as now, (221 22) (2n+2) [2n+1) goes to to faster than 12. 2 An nsa (227240) (en+ 2)/2n+1) 70 l'm 4 no (2n+2) (2011) = 0 and O o Radious of convergence is ! ... R Interval pl- of Convergence is (RR) ie (0o, as) the function Convergence f(n) C has the interval of de oven entine real line. (مه رو-)