Question:

Step 1: Identify the general term of the series. The given series is $S = \dfrac{x}{1-x^2} + \dfrac{x^2}{1-x^4} + \dfrac{x^4}{1-x^8} + \cdots$. Observing the pattern, the $n$-th term, $T_n$, can be written as: $$T_n = \dfrac{x^{2^{n-1}}}{1-x^{2^n}}$$ For example, for $n=1$, $T_1 = \dfrac{x^{2^0}}{1-x^{2^1}} = \dfrac{x}{1-x^2}$. For $n=2$, $T_2 = \dfrac{x^{2^1}}{1-x^{2^2}} = \dfrac{x^2}{1-x^4}$. Step 2: Rewrite the general term using an algebraic identity. We use the algebraic identity $\dfrac{a}{1-a^2} = \dfrac{1}{1-a} - \dfrac{1}{1-a^2}$. To verify this identity: $$\dfrac{1}{1-a} - \dfrac{1}{1-a^2} = \dfrac{1}{1-a} - \dfrac{1}{(1-a)(1+a)} = \dfrac{(1+a) - 1}{(1-a)(1+a)} = \dfrac{a}{(1-a)(1+a)} = \dfrac{a}{1-a^2}$$ Now, let $a = x^{2^{n-1}}$. Then $a^2 = (x^{2^{n-1}})^2 = x^{2^{n-1} \times 2} = x^{2^n}$. Applying this identity to the general term $T_n$: $$T_n = \dfrac{x^{2^{n-1}}}{1-x^{2^n}} = \dfrac{1}{1-x^{2^{n-1}}} - \dfrac{1}{1-x^{2^n}}$$ Step 3: Write out the first few terms of the series using the decomposed form. Using the rewritten form of $T_n$: For $n=1$: $T_1 = \dfrac{1}{1-x^{2^0}} - \dfrac{1}{1-x^{2^1}} = \dfrac{1}{1-x} - \dfrac{1}{1-x^2}$ For $n=2$: $T_2 = \dfrac{1}{1-x^{2^1}} - \dfrac{1}{1-x^{2^2}} = \dfrac{1}{1-x^2} - \dfrac{1}{1-x^4}$ For $n=3$: $T_3 = \dfrac{1}{1-x^{2^2}} - \dfrac{1}{1-x^{2^3}} = \dfrac{1}{1-x^4} - \dfrac{1}{1-x^8}$ The sum of the series to $N$ terms, $S_N$, is: $$S_N = \left(\dfrac{1}{1-x} - \dfrac{1}{1-x^2}\right) + \left(\dfrac{1}{1-x^2} - \dfrac{1}{1-x^4}\right) + \left(\dfrac{1}{1-x^4} - \dfrac{1}{1-x^8}\right) + \cdots + \left(\dfrac{1}{1-x^{2^{N-1}}} - \dfrac{1}{1-x^{2^N}}\right)$$ Step 4: Observe the telescoping cancellation and express the sum to infinity. This is a telescoping series, where all intermediate terms cancel out. $$S_N = \dfrac{1}{1-x} - \dfrac{1}{1-x^{2^N}}$$ To find the sum of the infinite series, we take the limit as $N \to \infty$: $$S = \lim_{N \to \infty} S_N = \lim_{N \to \infty} \left(\dfrac{1}{1-x} - \dfrac{1}{1-x^{2^N}}\right)$$ $$S = \dfrac{1}{1-x} - \lim_{N \to \infty} \dfrac{1}{1-x^{2^N}}$$ Step 5: Evaluate the limit. The problem states that $|x| < 1$. As $N \to \infty$, the exponent $2^N$ also approaches infinity. Since $|x| < 1$, we have $x^{2^N} \to 0$ as $N \to \infty$. Therefore, the limit term becomes: $$\lim_{N \to \infty} \dfrac{1}{1-x^{2^N}} = \dfrac{1}{1-0} = 1$$ Step 6: Calculate the final sum. Substitute the value of the limit back into the expression for $S$: $$S = \dfrac{1}{1-x} - 1$$ To simplify, find a common denominator: $$S = \dfrac{1}{1-x} - \dfrac{1-x}{1-x}$$ $$S = \dfrac{1 - (1-x)}{1-x}$$ $$S = \dfrac{1 - 1 + x}{1-x}$$ $$S = \dfrac{x}{1-x}$$ The sum of the series is $\dfrac{x}{1-x}$. The final answer is $\boxed{\dfrac{x}{1-x}}$.