1. **State the problem:** We want to analyze the infinite series $$\sum_{n=1}^\infty \left(\frac{7|x|n + 3}{6n + 3}\right)^{2n}$$ and determine for which values of $x$ it converges. 2. **Recall the root test for convergence:** For a series $\sum a_n$, the root test uses the limit $$L = \lim_{n \to \infty} \sqrt[n]{|a_n|}.$$ If $L < 1$, the series converges absolutely; if $L > 1$, it diverges; if $L = 1$, the test is inconclusive. 3. **Apply the root test:** Here, $$a_n = \left(\frac{7|x|n + 3}{6n + 3}\right)^{2n}.$$ Then, $$\sqrt[n]{|a_n|} = \sqrt[n]{\left(\frac{7|x|n + 3}{6n + 3}\right)^{2n}} = \left(\frac{7|x|n + 3}{6n + 3}\right)^2.$$ 4. **Simplify the expression inside the limit:** As $n \to \infty$, the dominant terms in numerator and denominator are $7|x|n$ and $6n$, so $$\lim_{n \to \infty} \left(\frac{7|x|n + 3}{6n + 3}\right)^2 = \left(\lim_{n \to \infty} \frac{7|x|n + 3}{6n + 3}\right)^2 = \left(\frac{7|x|}{6}\right)^2 = \frac{49|x|^2}{36}.$$ 5. **Set the root test condition for convergence:** We require $$\frac{49|x|^2}{36} < 1 \implies |x|^2 < \frac{36}{49} \implies |x| < \frac{6}{7}.$$ 6. **Conclusion:** The series converges absolutely for all $x$ such that $$|x| < \frac{6}{7}.$$