Question

12. For what values of r does the series (2n)!r" 22n(n!) converge absolutely? converge conditionally? diverge? n=1

Answer 1

Given that (2n)!** 22* (x!) (2n)! (2n+2)!**+1 Let a, = 2 ** (103 and auta – 220v? (n+1))* Ord - (2n+2)!***2 22* (nl)? am 22***(n+1)0)* (2n)! ** (2n+2)(2x +1) (2n)!x 224 (r!)? 22*** ((n +1)x!) (2n)! (2n+2)(2x+1)x 4(n+1) = (2+3)(2+4) (2+2)2+): - = x 41+1 Series ratio test the given series converges absolutely if x <1 Series ratio test the given series diverges if x >1 1 2 <1 then a converges IF 2 >1 then diverges I 2 = 1 then the test is inconclusive if x = 1 then § (2n)! . = 2*n! 22(n = lim =0 *+ 2n! (2n)! 2.4.6.-(2n) lim- =lim *** 22" (n) ** 22(n) conditionally convergent if x = 1 a, is decreasing sequence a, >0 for all n converges absolutely if x <1 diverges if x > 1 conditionally convergent if x = 1