I have to compute limn→∞nn(n!)2.
I tried say that this limit exists and it's l, so we have limn→∞nn(n!)2=L then I rewrited it as:
limn→∞(√nn√n!)2n then I used natural log over the whole expresion but didn't got into a nice place.
I don't know about Pi function or gamma function so therefore can't really use L'Hospital's rule.
Answer
By ratio test
(n+1)n+1((n+1)!)2(n!)2nn=1n+1(1+1n)n→0
then
limn→∞nn(n!)2=0
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