1. **State the problem:** Simplify the expression $$\frac{1}{\sqrt{2}+1}+\frac{1}{\sqrt{3}+\sqrt{2}}+\frac{1}{\sqrt{4}+\sqrt{3}}$$. 2. **Rationalize each denominator separately:** For the first fraction: $$\frac{1}{\sqrt{2}+1} \cdot \frac{\sqrt{2}-1}{\sqrt{2}-1} = \frac{\sqrt{2}-1}{(\sqrt{2})^2 - 1^2} = \frac{\sqrt{2}-1}{2 - 1} = \sqrt{2} - 1.$$ 3. For the second fraction: $$\frac{1}{\sqrt{3}+\sqrt{2}} \cdot \frac{\sqrt{3}-\sqrt{2}}{\sqrt{3}-\sqrt{2}} = \frac{\sqrt{3}-\sqrt{2}}{3 - 2} = \sqrt{3} - \sqrt{2}.$$ 4. For the third fraction: $$\frac{1}{\sqrt{4}+\sqrt{3}} \cdot \frac{\sqrt{4}-\sqrt{3}}{\sqrt{4}-\sqrt{3}} = \frac{2 - \sqrt{3}}{4 - 3} = 2 - \sqrt{3}.$$ 5. **Sum all simplified expressions:** $$ (\sqrt{2} - 1) + (\sqrt{3} - \sqrt{2}) + (2 - \sqrt{3}) = \sqrt{2} - 1 + \sqrt{3} - \sqrt{2} + 2 - \sqrt{3} = (\sqrt{2} - \sqrt{2}) + (\sqrt{3} - \sqrt{3}) + (-1 + 2) = 0 + 0 + 1 = 1.$$ 6. **Final answer:** $$1.$$