91Ó°ÊÓ

First, let's find the power series representation for \(\mathrm{e}^{x}\), which is: $$\mathrm{e}^{x} = \sum_{k=0}^{\infty}\frac{x^k}{k!}$$ Next, we need to find the power series representation for \(\frac{1}{1-x}\). This is a geometric series with a common ratio x: $$\frac{1}{1-x}=\sum_{k=0}^{\infty} x^k$$ Now, let us obtain the power series for the function \(f(x)\). To do that, we need to multiply both power series: $$f(x)=\frac{\mathrm{e}^{x}}{1-x} = \left(\sum_{k=0}^{\infty}\frac{x^k}{k!}\right)\left(\sum_{j=0}^{\infty} x^j\right)$$ To find the coefficient to \(x^n\), we will multiply the terms of the power x and sum the coefficients. In other words, we will use the Cauchy's product: $$a_n=\sum_{k=0}^{n}\frac{1}{k!}$$

To determine the range of x for which the power series converges, we can use the radius of convergence. Let's analyze the convergence of \(\frac{1}{1-x}\), which is a geometric series: The geometric series \(\sum_{k=0}^{\infty} x^k\) converges if and only if \(|x| < 1\). That means the power series converges for \(-1<x<1\). Now let's analyze the convergence of the power series of \(\mathrm{e}^{x}\), which is given by the ratio test: $$\lim_{k\to\infty}\left|\frac{a_{k+1}}{a_{k}}\right| = \lim_{k\to\infty}\left|\frac{\frac{x^{k+1}}{(k+1)!}}{\frac{x^k}{k!}}\right| = \lim_{k\to\infty}\left|\frac{k!}{(k+1)!}x\right| = \lim_{k\to\infty}\left|\frac{x}{k+1}\right|$$ Since the limit is 0 for every x, the power series of \(\mathrm{e}^{x}\) converges for every \(x\in \mathbb{R}\). As the function f(x) will be a product of both power series, we must consider both convergence ranges. So, since the power series converges for \(-1<x<1\), their product will converge for the same range. Thus, the power series representation of \(f(x)\) converges for \(x \in (-1,1)\).