limit 2^n*n!/n^n, n=infinity
The step-by-step solution converts to a log and then the answer really just begs the question. Is there a more detailed way of finding this limit?
Glad to be helpful!
It seems that this web site does not allow me to write -1 less than r less than 1.
This web site corrupts my formulas. I meant that when -1<r<1 the sequence goes to zero.
-1<r<1
Good point, in this case one can deduce the limit without knowing details of the factorial expansion at infinity.
Short answer is no. A detailed set of steps would require some serious understanding of how to recast the factorial behavior at infinity in terms of logs and exponentials.
Take the ratio of two consecutive terms:
a[n_] = 2^n n!/n^n; FullSimplify[a[n]/a[n + 1], n > 0]
You will see that a[n+1]/a[n] converges to 2/e, which is less than 1. You can also try
a[n+1]/a[n]
2/e
DiscretePlot[2^n n!/n^n, {n, 1, 20}]
