<s>[INST] How do I find the total area of total area of the red circles C_n,n=1,2,3 (Calculus II) 
I am very confused with this problem. I have been working on it for about an hour and still nothing. Help would be greatly appreciated!
 [/INST] One possible solution for this uses Decartes' Circle Theorem, which states that for four mutually externally tangent circles with radii $a,b,c,d$, then the following equality holds:
$$
2\left(\frac{1}{a^2}+\frac{1}{b^2}+\frac{1}{c^2}+\frac{1}{d^2}\right)=\left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}+\frac{1}{d}\right)^2
$$
Let $r_n$ be the radius of circle $n$. Use the distance formula between the center of $C_1$ and the center of one of the big circles to obtain
$$
(1-r_1)^2+1^2=(1+r_1)^2\Rightarrow r_1=\frac{1}{4}
$$
We claim by induction that $r_n=\frac{1}{2n(n+1)}$. The base case is trivial, and given $r_n$ is decreasing, it suffices to show that
$$
2(4(n-1)^2n^2+4n^2(n+1)^2+2)=(2(n-1)n+2n(n+1)+2)^2
$$
Or equivalently that
$$
(4n^2(n^2+1)+1)=(2n^2+1)^2
$$
which is true.
Now we want to find
$$
\frac{\pi}{4}\left(\displaystyle\sum_{n\ge 1} \frac{1}{n^2(n+1)^2}\right)
$$
But since
$$
\frac{1}{n^2(n+1)^2} = \frac{1}{n^2}+\frac{1}{(n+1)^2}+2\left(\frac{1}{n+1}-\frac{1}{n}\right)
$$
The answer is
$$
\frac{\pi}{4}\left(\frac{\pi^2}{3}-3\right)=\frac{\pi^3}{12}-\frac{3\pi}{4}
$$
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