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The question: (1) Prove that $\ln(n!) = (n+\frac{1}{2})\ln(n) - n + C + o(1)$, where $C - const$, o(1) - infinitely small.

(2) Using Valles' formula $\lim_{n\to\infty}\frac{1}{n}(\frac{(2n)!!}{(2n-1)!!})^2 = \pi$, prove that $C = \frac{\ln(2\pi)}{2}$.

(3) Derive Sterling's formula from your work ($n! \sim \sqrt{2\pi n}(\frac{n}{e})^n $).

What I have done:

I have proven ($1$) (by comparing $C$ for $n$ and $n-1$), and I believe I can approximate C by taking $n = 1$ and getting the following formula:

$\ln1 = (1+\frac{1}{2})\ln1 - 1 + C + o(1)$

$0 = 0-1+C+o(1)$

$C = 1 - o(1)$

But I don't think it fits the approximation of $\frac{\ln(2\pi)}{2}$ which is $\approx 0.919$, since $o(1)$ is supposed to be infinitely small.

I don't really know where to go from here, so, if you could, please point me in the right direction.

waterfalls
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    Have you tried (repeated) comparisons series/integral? E.g., as in this answer. (But you'll need to do it two more times, to get respectively the $\ln n$ and constant.) See also this answer, and more precisely the comments by Antonio Vargas, to see an illustration of the technique -- although for a different sum, this time. – Clement C. Feb 25 '16 at 13:06
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    Also, to explain why what you did does not yield the right value for $C$: $o(1)$ means "quantity that goes to $0$ when $n\to\infty$," but this asymptotic behavior may only kick in for large values of $n$. For $n=1$, this $o(1)$ term could literally be anything. – Clement C. Feb 25 '16 at 13:27
  • http://math.stackexchange.com/questions/94722/stirlings-formula-proof/1409131#1409131 – Jack D'Aurizio Feb 25 '16 at 13:30
  • @JackD'Aurizio But the exercise appears to ask to prove Striling's formula in a very specific way. – Clement C. Feb 25 '16 at 13:33
  • @Clement C. We haven't yet gotten to definite integrals yet, so I'm afraid I can't use the second method. I will look into the first one, thank you. And thanks for the clarification on o(1), I didn't know that. – waterfalls Feb 25 '16 at 13:33
  • @MaximGrigorovich You say you have proven $(1)$ -- how did you, exactly? (this may help understanding also what tools you know and are allowed to use.) – Clement C. Feb 25 '16 at 14:40
  • @ClementC. Sorry for the late reply. $$\ln(n!) = (n+\frac{1}{2})\ln(n)-n+C+o(1)$$

    $$\ln((n-1)!) = (n-1+\frac{1}{2})\ln(n-1)-n+1+C+o(1)$$

    Get C values from these 2 and compare them.

    $$(n+\frac{1}{2})\ln(n) - n - \ln(n!) + o(1) = (n-\frac{1}{2})\ln(n-1)-n+1-\ln((n-1)!) + o(1)$$

    ...

    $$(n-\frac{1}{2})\ln(\frac{n}{n-1}) = 1 + o(1)$$

    $$\frac{n}{n-1}^{(n-\frac{1}{2})} = e + o(1) $$ (Not really sure it would be e + o(1) but it should be something close to that)

    $$(1+ \frac{1}{n-1})^{(n-\frac{1}{2})} = e + o(1)$$ which is true since i know that $(1+\frac{1}{n-1})^n$ approaches e

     – waterfalls Feb 26 '16 at 13:03
  • @MaximGrigorovich OK. So technically, you did not prove the statement: you proved it was consistent, right? – Clement C. Feb 26 '16 at 14:57

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