Squaring an arbitrary summation? I'm trying to find a recurrence relation for the coefficients for the Maclaurin series for $\tan(x)$ by substituting $y=\sum_{k=0}^{\infty}C_{2k+1}x^{2k+1}$ into the differential equation $y'=1+y^2$. This is because $\tan(x)$ is the solution to the initial value problem for the aforementioned DE with the initial condition $y(0)=0$; this is also where the form $\sum_{k=0}^{\infty}C_{2k+1}x^{2k+1}$ comes from (the fact that $\tan(x)$ is an odd function and that $y(0)=0$ which implies $C_0=0$). But I have no clue how to work "around" the expression $y^2=\big(\sum_{n=1}^{\infty}C_{2k+1}x^{2k+1}\big)^2$. How can I find a recurrence relation with an infinite squared summation? Any help is appreciated, thank you.

Take the Cauchy product:
$$\begin{align*}
\left(\sum_{k\ge 0}C_{2k+1}x^{2k+1}\right)^2&=x^2\left(\sum_{k\ge 0}C_{2k+1}x^{2k}\right)^2\\\\
&=x^2\sum_{k\ge 0}D_{2k}x^{2k}\;,
\end{align*}$$
where $$D_{2k}=\sum_{i=0}^kC_{2i+1}C_{2(k-i)+1}\;.$$ Thus, the differential equation becomes
$$\begin{align*}
\sum_{k\ge 0}C_{2k+1}(2k+1)x^{2k}&=1+x^2\sum_{k\ge 0}\sum_{i=0}^kC_{2i+1}C_{2(k-i)+1}x^{2k}\\
&=1+\sum_{k\ge 1}\sum_{i=0}^{k-1}C_{2i+1}C_{2(k-i)-1}x^{2k}\;,
\end{align*}$$
and we have $C_1=1$ and $$C_{2k+1}=\frac1{2k+1}\sum_{i=0}^{k-1}C_{2i+1}C_{2(k-i)-1}$$ for $k\ge 1$. 
E.g., 
$$\begin{align*}
C_3&=\frac13 C_1^2=\frac13\;,\\
C_5&=\frac15(2C_1C_3)=\frac2{15}\;,\text{ and}\\
C_7&=\frac17(2C_1C_5+C_3^2)=\frac{17}{315}\;.
\end{align*}$$
