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Editorial Team
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Asked: 2026-05-13T11:35:04+00:00

I am to show that log( n !) = Θ( n ·log( n ))

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I am to show that log(n!) = Θ(n·log(n)).

A hint was given that I should show the upper bound with nn and show the lower bound with (n/2)(n/2). This does not seem all that intuitive to me. Why would that be the case? I can definitely see how to convert nn to n·log(n) (i.e. log both sides of an equation), but that’s kind of working backwards.

What would be the correct approach to tackle this problem? Should I draw the recursion tree? There is nothing recursive about this, so that doesn’t seem like a likely approach..

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1 Answer

  1. Editorial Team

    Remember that 

    log(n!) = log(1) + log(2) + ... + log(n-1) + log(n)

    You can get the upper bound by 

    log(1) + log(2) + ... + log(n) <= log(n) + log(n) + ... + log(n)
                                = n*log(n)

    And you can get the lower bound by doing a similar thing after throwing away the first half of the sum:

    log(1) + ... + log(n/2) + ... + log(n) >= log(n/2) + ... + log(n) 
                                       = log(n/2) + log(n/2+1) + ... + log(n-1) + log(n)
                                       >= log(n/2) + ... + log(n/2)
                                        = n/2 * log(n/2)
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