1. **State the problem:** Show that $$\frac{1}{2}(e - \frac{1}{e}) = 1 + \frac{1}{3!} + \frac{1}{5!} + \ldots$$ 2. **Recall the known series expansion:** The hyperbolic sine function $$\sinh x$$ is defined as: $$\sinh x = \frac{e^x - e^{-x}}{2} = x + \frac{x^3}{3!} + \frac{x^5}{5!} + \frac{x^7}{7!} + \ldots$$ 3. **Apply the formula for $$x=1$$:** Substitute $$x=1$$ into the series: $$\sinh 1 = \frac{e^1 - e^{-1}}{2} = 1 + \frac{1^3}{3!} + \frac{1^5}{5!} + \frac{1^7}{7!} + \ldots = 1 + \frac{1}{3!} + \frac{1}{5!} + \ldots$$ 4. **Compare with the left-hand side (LHS):** The LHS is exactly $$\frac{1}{2}(e - \frac{1}{e})$$ which equals $$\sinh 1$$. 5. **Conclusion:** Therefore, the equality $$\frac{1}{2}(e - \frac{1}{e}) = 1 + \frac{1}{3!} + \frac{1}{5!} + \ldots$$ holds true by the definition and series expansion of the hyperbolic sine function.