Question

Sigma Sum n=1,INFINITY[((n!)^2)/(kn!)]

For which positive integers k is the series convergent?...I am stumped...

Answer 1

\[\sum_{n=1}^{\infty} \frac{n!^2}{kn!}\]
I really dont remember my convergence tests... but it seems like the numerator will approach infinity much quicker than the denominator for all k not equal to 0. which would cause it to be perminately divergent except for k=0 where it is undefined.

Answer 2

I thought k= infinity was right as well. it is wrong. The answer is something like k\[\le\]???

Answer 3

Oops reverse that....

Answer 4

i thought k = 0 ....not sure why it plopped infinity in there.

Answer 5

Are you sure that the equation isn't \[\sum_{n=1}^{\infty} {{(n!)^{2}}\over{(kn)!}} \]

Answer 6

Yeah, thats the equation

Answer 7

Any ideas?

Answer 8

Well it definitely converges for some k...
Let me think about it for a bit

Answer 9

thanks brah

Answer 10

Well I'm pretty sure it converges for k >= 2

Answer 11

sure does! THanks

Answer 12

Whats the logic?

Answer 13

So, n! grows a lot faster than n^2

Answer 14

so (kn)! is applying the factorial after you multiply by a constant where (n!)^2 evaluates n! then applies the square

Answer 15

The distance between consecutive factorial terms grows a lot faster than squaring terms

Answer 16

But if k >= 2 then we are at least doubling that distance before we even apply the factorial

Answer 17

10! << 20!

Answer 18

(10!)^2 = (10!)(10!) < (20!) = (20*19*18*17*16*15*14*13*12*11)*10!

Answer 19

hmmm well thanks! Gotta think about it more.

Answer 20

Use the ratio test. The sum of an infinite series x_n converges if
limit |x_{n+1} / x_n | < 1 as n --> infinity.