# How to compare convergence of two equations for the arctangent function?

I want to compare two equations for the arctangent function: $$\tag{1}\arctan\left(x\right)=\sum\limits_{n=0}^{\infty}{\frac{{2^{2n}}{\left(n! \right)}^{2}}{\left(2n+1\right)!}\frac{x^{2n+1}}{{\left(1+x^2\right)}^{n+1}}}$$ and $$\tag{2}\arctan\left(x\right)=i\sum\limits_{n=1}^{\infty}{\frac{1}{2n-1}}\left(\frac{1}{\left(1+2i/x\right)^{2n-1}}-\frac{1}{\left(1-2i/x\right)^{2n-1}}\right).$$

Which of them is faster in convergence?

• Simple numerical experiments suggest that the second sum converges faster. – Michael Renardy Jun 6 '17 at 18:08
• @ Michael Thank you! But how much faster? How to compare these equations for convergence or accuracy? – Freedom Math Jun 6 '17 at 18:17

If $x > 0$, let $r = 1/|1 + 2 i/x| = 1/\sqrt{1 + 4/x^2}$ and the $n$'th term on the right of the second sum is dominated by $2/((2n-1)r^{2n-1})$.
On the other hand, for large $n$ the $n$'th term in the first sum is approximately $\sqrt{\pi/(4n)} (x/(1+x^2)) (x^2/(1+x^2))^n$. Since $$\frac{x^2}{1+x^2} > \frac{1}{1+4/x^2} = \frac{x^2}{4 + x^2}$$ for all $x > 0$ we find that the $n$'th term in the second sum will go to $0$ asymptotically faster.
Here is an animation showing the absolute values of the errors in the $N$'th partial sums for $N$ from $1$ to $50$ and $x$ from $0$ to $3$. The red curve is for the first series and the blue curve is for the second.
• @ Robert Thank you! Your description and error graphs clearly show that the convergence of the second equation is more rapid, especially at $x<4$ as I understand. – Freedom Math Jun 9 '17 at 15:19