Question

Use the power series $$\frac{1}{1+x}=\sum_{n=0}^{\infty}(-1)^{n} x^{n}$$ to determine a power series, centered at 0 , for the function. Identify the interval of convergence. $$ f(x)=-\frac{1}{(x+1)^{2}}=\frac{d}{d x}\left[\frac{1}{x+1}\right] $$

Short answer

The required power series for the function \(f(x)\) is \(\sum_{n=1}^{\infty}(n)(-1)^{n}x^{n-1}\). The interval of convergence is \( -1<x<1 \), still excluding the endpoints as the series does not converge at \(x=-1\) and \(x=1\).

Step-by-step solution

Identify Given Power Series

The given power series is \( \frac{1}{1+x}=\sum_{n=0}^{\infty}(-1)^{n}x^{n} \). The function \(f(x)\) can be expressed as the derivative of the given power series, which means that we need to take the derivative of the right side of the equation.

Deriving the Power Series

By taking the derivative of \(\sum_{n=0}^{\infty}(-1)^{n}x^{n}\), we get the following series: \(\sum_{n=1}^{\infty}(n)(-1)^{n}x^{n-1}\). Notice that the result looks almost like the original series, but multiplied by \( n \) and the power of \(x\) decreases by 1.

Calculate Interval of Convergence

For the series \(\sum_{n=0}^{\infty}(-1)^{n}x^{n}\), the interval of convergence is \( -1<x<1 \). After differentiating, the interval of convergence remains the same, hence, the interval of convergence of the series for \(f(x)\) is also \( -1<x<1 \). Though, keep in mind that the convergence at the endpoints of this interval needs to be checked separately.

Checking Endpoint Convergence

When \(x=-1\), the series becomes \(\sum_{n=1}^{\infty}(n)(-1)^{n-1}\), which does not converge. Similarly, when \(x=1\), the series becomes \(\sum_{n=1}^{\infty}(n)(-1)^{n}\), which also does not converge.