2.1 Given: \( x+\frac{1}{2} x^{2}+\frac{1}{4} x^{3}+\frac{1}{8} x^{4}+\ldots \) For which values of \( x \) will the series converge? Show working details.

The given series is an infinite series expressed as \( x+\frac{1}{2} x^{2}+\frac{1}{4} x^{3}+\frac{1}{8} x^{4}+\ldots \). We can see that it can be rewritten in summation notation as: \[ \sum_{n=1}^{\infty} \frac{1}{2^{n-1}} x^n \] This is a power series where each term is of the form \( a_n x^n \) with \( a_n = \frac{1}{2^{n-1}} \). We can determine the convergence of this series using the ratio test. Using the ratio test, we calculate the limit: \[ L = \lim_{n \to \infty} \left| \frac{a_{n+1}}{a_n} \right| \] Here, \( a_{n+1} = \frac{1}{2^n} x^{n+1} \) and \( a_n = \frac{1}{2^{n-1}} x^n \). Therefore: \[ \frac{a_{n+1}}{a_n} = \frac{\frac{1}{2^n} x^{n+1}}{\frac{1}{2^{n-1}} x^n} = \frac{x}{2} \] We take the limit: \[ L = \lim_{n \to \infty} \left| \frac{x}{2} \right| = \left| \frac{x}{2} \right| \] For the series to converge, we need \( L < 1 \): \[ \left| \frac{x}{2} \right| < 1 \] This implies: \[ |x| < 2 \] Thus, the series converges for values of \( x \) in the interval: \[ (-2, 2) \] In conclusion, the series converges for \( x \) in the range \( (-2, 2) \).