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.
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}]
Hi Gianluca,
Thank you for taking the time to reply. The limit that I am trying to test for convergence is the limit of a sequence. The ratio test which you suggest, is used to determine the convergence of a series, not a sequence.
If you cut and paste my original command limit 2^n*n!/n^n, n=infinity into Wolfram Alpha you will see what I mean.
Regards, Brad
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