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Exponentiating, we getn! =

√ 2π nn+

1

2 e−nean,

where

an  =  1

12n −  1

3

  ∞n

B̃3(x)

x3  dx.

As we saw in an earlier exercise, |B̃3(x)| ≤ √ 336   for all  x ∈ R. Since

1

3

  ∞n

B̃3(x)

x3  dx

≤  1

3

  ∞n

|B̃3(x)|x3

  dx ≤√ 

3

108

  ∞n

1

x3  dx =

√ 3

216n2 <

  1

100n2.

This tells us that an   differs from   112n  be less than   1

100n2 . It tells us that  ean is roughly  e1/(12n) ≈1 +   1

12n, so that the true value of  n! is approximately   100

12n% bigger than its approximate value√ 

2πnn+1/2e−n.

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