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I am working through a proof of Stirling's Formula in Feller's An Introduction to Probability Theory and it's Applications and am stuck at equation 9.10, where he make a comparison with a geometric series. For full context, he states:

And using the expansion we get: $$d_n - d_{n+1} = \frac{1}{3(2n+1)^2} + \frac{1}{5(2n+1)^4}+ \dots\tag{9.9}\label{9.9}$$ By comparison of the right side with a geometric series with ratio $(2n+1)^{-2}$ one sees that: $$0 < d_n - d_{n+1} < \frac{1}{3[(2n+1)^2 - 1]} = \frac{1}{12n} - \frac{1}{12(n+1)}\tag{9.10}\label{9.10}$$

I am struggling to make the jump from equation 9.9 to 9.10. The geometric series with ratio $(2n+1)^{-2}$ would be (based on wikipedia article):

$$\frac{1}{(2n+1)^2} + \frac{1}{(2n+1)^4} + \frac{1}{(2n+1)^6} + \dots = \frac{1}{1 - \frac{1}{(2n+1)^2}}$$

Which leaves me trying to make a comparison between:

$$ d_n - d_{n+1} = \frac{1}{3(2n+1)^2} + \frac{1}{5(2n+1)^4}+ \dots < \frac{1}{1 - \frac{1}{(2n+1)^2}} $$

Which intuitively makes sense seeing as all terms on the left hand side have an additional factor in the denominator, ensuring it is less than the right hand side. However, I am still stuck trying to figure out how Feller arrives at equation 9.10.

Any help or input on where I am going wrong is greatly appreciated.

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    $\begingroup$ The RHS should be $ \frac{1}{(2n+1)^2}\frac{1}{1 - \frac{1}{(2n+1)^2}}$ because your geometric series does not begin at $1$ and its ratio is $ \frac{1}{(2n+1)^2}$. $\endgroup$
    – Jean Marie
    May 12, 2020 at 13:58
  • $\begingroup$ Appreciate the response @JeanMarie. I have a follow up question though-how do you determine if the geometric series should start at $1$ (as it does in the wikipedia article), or $\frac{1}{(2n+1)^2}$, as I had it start in my question? In the textbook Feller simply states "By comparison of the right side with a geometric series with a ratio of $(2n+1)^{-2}$"-this says nothing about the initial value, so how do you determine that? Thanks! $\endgroup$
    – ndake11
    May 12, 2020 at 14:08
  • $\begingroup$ I was only commenting on your penultimate formula. $\endgroup$
    – Jean Marie
    May 12, 2020 at 14:16

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Your formula for the sum of a geometric series is slightly off. First of all your formula would be the sum of a geometric series with ratio $(2n+1)^{-1}$, not $(2n+1)^{-2}$. More importantly however, the expression $\frac1{1-r}$ is the sum of $1+r+r^2+...$, however in the current case we're lacking the first term, we instead have the "decapitated" geometric series $r+r^2+r^3...$ with sum

$$S=\frac{1}{1-r}-1=\frac{r}{1-r}=\frac1{(2n+1)^2-1}=\frac1{4n^2+4n}=\frac1{4n(n+1)}$$

That last fraction can be rewritten using partial fractions

$$S=\frac{1}{4n}+\frac{-1}{4(n+1)}$$

Each term in the original series is a term from the geometric series multiplied by a value less than or equal to $\frac13$, therefore the sum of the series is at most

$$\frac13S=\frac{1}{12n}-\frac{1}{12(n+1)}$$

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  • $\begingroup$ That was very helpful and super clear-thank you so much! I have a brief follow up question now that that is settled. Feller proceeds to say that: "From 9.9 we conclude that the sequence {$d_n$} is decreasing, while 9.10 shows that the sequence {$d_n - (12n)^{-1}$} is increasing. It follows that the finite limit $C = lim d_n$ exists". Can you explain why the equations 9.9 and 9.10 in junction allow us to say the limit exists? Why is 9.10 needed to show this? $\endgroup$
    – ndake11
    May 12, 2020 at 15:22
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    $\begingroup$ (9.9) shows that $d_n$ is decreasing because the right hand side is positive. Now we just need to exclude the possibility $d_n\to-\infty$. Let $a_n=\frac1{12n}$. The idea here seems to be that if $d_n-a_n$ is increasing, then $d_n\to-\infty$ is impossible, since if $d_n\to-\infty$, then since $a_n$ is bounded below, $d_n-a_n$ would have to decrease at least some of the time. $\endgroup$
    – Jack M
    May 12, 2020 at 19:26
  • $\begingroup$ One final question-Feller proceeds to then say that 9.10 has a companion inequality in the reverse direction; from 9.9 and 9.10 it is obvious that $d_n - d_{n+1} > \frac{1}{3(2n+1)^2} > \frac{1}{12n + 1} - \frac{1}{12(n+1) +1}$. The first part of this inequality is of course straightforward based on 9.9, but I am not able to figure out where the right hand side is coming from. Any help is much appreciated! $\endgroup$
    – ndake11
    May 13, 2020 at 15:21
  • $\begingroup$ @ndake11 I don't see where that's coming from. You might consider asking a new question if you can't work it out. $\endgroup$
    – Jack M
    May 15, 2020 at 11:51

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