Download Proof of the stirling`s formula

Stirling’s formula
Shatha sattar: [email protected]
October 29, 2014
An important formula in applied mathematics as well as in probability is the Stirling's formula
known as
where
is used to indicate that the ratio of the two sides goes to 1 as n goes to
words, we have
. In other
Or
Proof of the stirling’s formula
first take the log of n! to get
Since the log function is increasing on the interval
, we get
For 𝑛 ≥ 1 . Add the above inequalities, with = 1,2, … . 𝑁 , we get
Though the first integral is improper, it is easy to show that in fact it is convergent. Using the
antiderivative of log( 𝑥) ( being x log( 𝑥) − 𝑥), we get
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Next, set
We have
Easy algebraic manipulation gives
Using the Taylor expansion
for -1 < t < 1, we get
This implies
We recognize a geometric series. Therefore we have
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From this we get
1. the sequence {𝑑𝑛 } is decreasing;
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2. the sequence {𝑑𝑛 − 12𝑛} is increasing.
This will imply that {𝑑𝑛 } converges to a number C with
and that C > d1 - 1/12 = 1 - 1/12 = 11/12. Taking the exponential of dn, we get
The final step in the proof if to show that 𝑒 𝑐 = √2𝜋 . This will be done via Wallis
formula (and Wallis integrals). Indeed, recall the limit
Rewriting this formula, we get
Playing with the numbers, we get
Using the above formula
We get
Easy algebra gives
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since we are dealing with constants, we get in fact 𝑒 𝑐 = √2𝜋 This completes the proof of
the Stirling's formula.
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