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Need to calculate a limit. Step-by-step solution doesn't give much guidance

Posted 9 years ago

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?

POSTED BY: Brad Cooper
9 Replies

Glad to be helpful!

POSTED BY: Gianluca Gorni

It seems that this web site does not allow me to write -1 less than r less than 1.

POSTED BY: Gianluca Gorni

This web site corrupts my formulas. I meant that when -1<r<1 the sequence goes to zero.

POSTED BY: Gianluca Gorni

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.

POSTED BY: Gianluca Gorni

Good point, in this case one can deduce the limit without knowing details of the factorial expansion at infinity.

POSTED BY: Daniel Lichtblau
POSTED BY: Daniel Lichtblau

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

DiscretePlot[2^n n!/n^n, {n, 1, 20}]
POSTED BY: Gianluca Gorni
Posted 9 years ago
POSTED BY: Brad Cooper
Posted 9 years ago

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

POSTED BY: Brad Cooper
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