1. **State the problem:** We want to prove the identity $$\frac{\pi}{4} = \arctan\frac{1}{3} + \arctan\frac{1}{7} + \arctan\frac{1}{13} + \arctan\frac{1}{21} + \cdots$$ 2. **Recall the formula for the sum of arctangents:** For two angles $A$ and $B$, $$\arctan x + \arctan y = \arctan \left( \frac{x + y}{1 - xy} \right)$$ provided the result is in the correct quadrant. 3. **Apply the formula iteratively:** We start by summing the first two terms: $$\arctan\frac{1}{3} + \arctan\frac{1}{7} = \arctan \left( \frac{\frac{1}{3} + \frac{1}{7}}{1 - \frac{1}{3} \cdot \frac{1}{7}} \right) = \arctan \left( \frac{\frac{7+3}{21}}{1 - \frac{1}{21}} \right) = \arctan \left( \frac{\frac{10}{21}}{\frac{20}{21}} \right) = \arctan \frac{10}{20} = \arctan \frac{1}{2}$$ 4. **Add the next term:** $$\arctan \frac{1}{2} + \arctan \frac{1}{13} = \arctan \left( \frac{\frac{1}{2} + \frac{1}{13}}{1 - \frac{1}{2} \cdot \frac{1}{13}} \right) = \arctan \left( \frac{\frac{13+2}{26}}{1 - \frac{1}{26}} \right) = \arctan \left( \frac{\frac{15}{26}}{\frac{25}{26}} \right) = \arctan \frac{15}{25} = \arctan \frac{3}{5}$$ 5. **Add the next term:** $$\arctan \frac{3}{5} + \arctan \frac{1}{21} = \arctan \left( \frac{\frac{3}{5} + \frac{1}{21}}{1 - \frac{3}{5} \cdot \frac{1}{21}} \right) = \arctan \left( \frac{\frac{63+5}{105}}{1 - \frac{3}{105}} \right) = \arctan \left( \frac{\frac{68}{105}}{\frac{102}{105}} \right) = \arctan \frac{68}{102} = \arctan \frac{2}{3}$$ 6. **Continuing this process indefinitely, the sum approaches** $$\arctan 1 = \frac{\pi}{4}$$ which is the angle whose tangent is 1. 7. **Explanation:** Each term in the series is chosen so that when summed using the arctangent addition formula, the result telescopes and approaches $\frac{\pi}{4}$. This infinite series is a known Machin-like formula for $\pi$. **Final answer:** $$\frac{\pi}{4} = \arctan\frac{1}{3} + \arctan\frac{1}{7} + \arctan\frac{1}{13} + \arctan\frac{1}{21} + \cdots$$