calculus and convergence question?

Question

If "k" is a positive integer, find the radius of convergence of the series,

summation notation symbol (with n=0 on bottom and infinity on top) of ((n!)^k/(kn)!) x x^n

show work/steps plz..thanks

Answer 1

According to the ratio test the series will converge if in the limit as n goes to infinity the ratio of the absolute value of the (n+1)th term of the sequence over the nth term goes to a limit less than 1.

lim n goes to infinity of |[(((n+1)!)^k/(k(n+1))!)x x^(n+1)]/[((n!)^k/(kn)!) x x^n]|

= |[(((n+1)^k)(n!)^k/(kn+k)!)x^2 x^n]/[((n!)^k/(kn)!) x x^n]|

= {[(n+1)^k (kn)!]/[(kn+k)!]}|x|

= |x|(n+1)^k/[(kn+1)...(kn+k)]

If we multiply numerator and denominator by 1/n^k we get,

= |x|(1+1/n)^k/[(k+1/n)...(k+k/n)]

In the limit as n approaches infinity we get |x|/k^k. That will be less than 1 when

|x| < k^k.

Answer 2

You know, math never made sense to me. See 2 + 2 = 4, and it will always be 4, so why the heck do people make it complicated? Who gives a rat's *** weather the square root of 4 is 16 or whatever it is? Who decided that x = 2+ z - c+44-33 = 12874602459946879847958712936598958. It is not like you need this stuff on a daily basis. As long as you know how to multiply, subtract, divide and add that is all you need, the whole x+z+a+x thing is crap, it is just crap!!!!

Answer 3

take abs val and use ratio test
a_n+1/a_n = (n+1)^k/(kn+1)*(kn+2)******(kn+k) IxI
The denominator is a polynome degree k and first term
k^k so the limit is IxI/k^k <1
so if IxI