If $f(x) = \lim_{n \to \infty} \left[2x + 4x^3 + ... + 2nx^{2n-1}\right]$ $(0 < x < 1)$ then $\int f(x)dx$ is equal to:
Step-by-Step Solution
Key Concept: The geometric series sum formula combined with limits allows us to find the limiting function, and the result is integrated using standard techniques.
Given $g_n(x) = 1 + x^2 + x^4 + \cdots + x^{2n} = \frac{x^{2n+2}-1}{x^2-1}$, we compute $g_n'(x) = \frac{2x(nx^{2n+2}-(n+1)x^{2n}+1)}{(x^2-1)^2}$. For $0 < x < 1$, $f(x) = \lim_{n \to \infty} h_n(x) = \frac{2x}{(x^2-1)^2}$. Integration yields $\int f(x) dx = -\frac{1}{x^2-1} + C$.
Correct Answer: 4