- #1
joe_cool2
- 24
- 0
I am to find whether the sum of (n!)/(n^n) converges or diverges. I tried both the limit comparison test, and a regular comparison test. (These are the only types of tests I am allowed to use.) So I tried several approaches:
Approach #1: (n!)/(n^n) > 1/(n^n)
Normally we use a setup like this to prove something with a p-series. However, the expression on the left side of the inequality isn't a p-series.
Approach #2: (n!)/(n^n) < n!
While this expression is true, it is not useful because the formula for the series is less than, not greater than, the series that is known to diverge.
Approach #3 (Limit comparison): an = (n!)/(n^n) ; bn = 1/(n^n)
an/bn = n!
The limit here is, unfortunately, infinite, and I have to stop here.
What other approach can I take that would result in more success?
Approach #1: (n!)/(n^n) > 1/(n^n)
Normally we use a setup like this to prove something with a p-series. However, the expression on the left side of the inequality isn't a p-series.
Approach #2: (n!)/(n^n) < n!
While this expression is true, it is not useful because the formula for the series is less than, not greater than, the series that is known to diverge.
Approach #3 (Limit comparison): an = (n!)/(n^n) ; bn = 1/(n^n)
an/bn = n!
The limit here is, unfortunately, infinite, and I have to stop here.
What other approach can I take that would result in more success?