1. **State the problem:** We want to determine the convergence of the infinite series $$\frac{x}{1 \cdot 2} + \frac{x^2}{3 \cdot 4} + \frac{x^3}{5 \cdot 6} + \cdots$$ where $x > 0$. 2. **Identify the general term:** The $n$-th term of the series can be written as $$a_n = \frac{x^n}{(2n-1)(2n)}$$ for $n = 1, 2, 3, \ldots$. 3. **Analyze the denominator:** Note that $$(2n-1)(2n) = 4n^2 - 2n$$ which grows approximately like $4n^2$ as $n$ becomes large. 4. **Apply the root test or ratio test:** Let's use the root test for convergence. Calculate $$\lim_{n \to \infty} \sqrt[n]{|a_n|} = \lim_{n \to \infty} \sqrt[n]{\frac{x^n}{(2n-1)(2n)}} = \lim_{n \to \infty} \frac{\sqrt[n]{x^n}}{\sqrt[n]{(2n-1)(2n)}} = \lim_{n \to \infty} \frac{x}{\sqrt[n]{(2n-1)(2n)}}$$ Since $(2n-1)(2n)$ grows without bound, its $n$-th root tends to 1: $$\lim_{n \to \infty} \sqrt[n]{(2n-1)(2n)} = 1$$ Therefore, $$\lim_{n \to \infty} \sqrt[n]{|a_n|} = x$$ 5. **Conclude convergence:** The root test states that the series converges if this limit is less than 1, diverges if greater than 1, and is inconclusive if equal to 1. Since $x > 0$, the series converges if and only if $$x < 1$$ and diverges if $$x > 1$$ 6. **Check the boundary case $x=1$:** When $x=1$, the series becomes $$\sum_{n=1}^\infty \frac{1}{(2n-1)(2n)}$$ We can use partial fraction decomposition: $$\frac{1}{(2n-1)(2n)} = \frac{A}{2n-1} + \frac{B}{2n}$$ Solving for $A$ and $B$: $$1 = A(2n) + B(2n-1)$$ Setting $n=\frac{1}{2}$ gives $1 = A(1) + B(0) \Rightarrow A=1$. Setting $n=0$ gives $1 = A(0) + B(-1) \Rightarrow B = -1$. So, $$\frac{1}{(2n-1)(2n)} = \frac{1}{2n-1} - \frac{1}{2n}$$ The series becomes $$\sum_{n=1}^\infty \left( \frac{1}{2n-1} - \frac{1}{2n} \right)$$ This is a telescoping series that converges to $\ln(2)$. **Final answer:** - The series converges for $0 < x \leq 1$. - The series diverges for $x > 1$.