Question

Determine whether the series converges absolutely, converges conditionally, or diverges

Answer 1

\[\sum_{n=1}^{\infty} -1^k {(k!)^3 \over (3k)!}\]

Answer 2

I got (1/27) which is less 1 so converges absolutely?

Can anyone confirm this answer?

Answer 3

maybe converges conditionally, not sure though... calculating it..

Answer 4

are you asking this\[\sum_{n=1}^{\infty} (-1)^k {(k!)^3 \over (3k)!}\]

Answer 5

is it alternate series ?

Answer 6

\[ \frac{(k+1)!^3}{(3k + 3)!} \frac{3k!}{k!^3} = \frac{k+1}{3!}\]If I did not make any careless mistake. This limit does not seem to converge as k -> infinity, so it abs. diverges?

Answer 7

http://www.wolframalpha.com/input/?i=limit+k+-%3E+infinity+%28%28%28%28k%2B1%29%21%29%5E3%29%2F%283%28k%2B1%29%21%29+*+%28%283k%29%21%2F%28k%21%29%5E3%29

Answer 8

@cinar yeah it is

@bmp why is the denominator 3! ? not (3k+3)(3k+2)(3k+1)?

Answer 9

I missed the parenthesis there, it should be 3(k+1)!, which is 3(k+1). But I was careless with my work up there. The series appear to diverge absolutely, but do the limit more carefully than what I did.

Answer 10

I guess you can use ratio test..