Divergence of 2^(n^2)/n!

1. Nov 9, 2012

peripatein

Hi,
How may I show that 2^(n^2)/n! converges to infinity?

2. Nov 9, 2012

Dick

That is SO divergent you can afford to be pretty sloppy. 2^(n^2)/n!>2^(n^2)/n^n, right? So show 2^(n^2)/n^n diverges. Hint: look at the log.

3. Nov 9, 2012

peripatein

Thank you, Dick!

4. Nov 9, 2012

peripatein

Would it be correct to say that if for sequences a_n and b_n, lim a_n = infinity and |b_n|< c < infinity, then lim|a_n*b_n| = infinity?
(I think it should be correct, as we may infer that lim |bn| = c and then the limit of the product of a_n and b_n would yield c*infinity which is always infinity.)

5. Nov 9, 2012

Dick

If you mean lim |b_n|=c with c>0, then sure. If c=0, then you need to think more.

Last edited: Nov 9, 2012