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
It seems that Wolfram|Alpha misunderstood your syntax. Try with
Limit of 2^n*n!/n^n as n goes to infinity
As for the ratio test, there is a version for sequences. Suppose that r=lim a[n+1]/a[n]. If -1<r<1 then a[n] goes to zero. If r>1 then a[n] goes to Infinity.
r=lim a[n+1]/a[n]
a[n]
r>1
Good point, in this case one can deduce the limit without knowing details of the factorial expansion at infinity.
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}]
Hi Gianluca,
Brilliant!! I never knew that there is a ratio test for sequences. Your reply to me prompted me to go and search for it and I found it documented here...
http://math.duke.edu/~cbray/Stanford/2003-2004/Math%2042/seqconv.pdf
As you say, this makes calculation of the limit of the sequence I am interested in beautifully simple. Thank you very much.
Cheers, Brad