Geometric series in proof of Stirling's Formula 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. 

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