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Iosif Pinelis
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More or less straightforward calculations show that $$E_n=\frac1{(2 n-1)!}\sum _{k=0}^n k! (2 n-k-1)! \binom{2 n-k+1}{k}.$$ This should be rather easy to analyze using Stirling's formula.

Iosif Pinelis
  • 127.7k
  • 8
  • 107
  • 229