Problem 64 - Calculus Early Transcendentals

Question

Identify the functions represented by the following power series. $$\sum_{k=2}^{\infty} \frac{x^{k}}{k(k-1)}$$

Step-by-Step Solution

Answer

Answer: Within its interval of convergence, the given power series converges to the function: $$f(x) = x\ln(1-x) - x + C_1x + C_2$$ where \(C_1\) and \(C_2\) are constants of integration.

Step 1: Rewrite as a derivative or integral of a known power series

First, let's notice that we can rewrite the given power series as follows: $$\sum_{k=2}^{\infty} \frac{x^{k}}{k(k-1)} = \sum_{k=2}^{\infty} \frac{1}{k-1} \cdot \frac{x^{k}}{k} = \sum_{k=2}^{\infty} \frac{1}{k-1} \cdot x^{k} \cdot \frac{1}{k}$$ Now, we can rewrite the sum by changing the index and noticing a pattern from k=2 to k=n: $$\sum_{n=1}^{\infty} \frac{1}{n} \cdot x^{n+1} \cdot \frac{1}{n+1}$$ Now we can notice that it can be the second derivative of the geometric series, with respect to x (without first term of the sum): $$\sum_{n=1}^{\infty} \frac{1}{n} \cdot x^{n+1} \cdot \frac{1}{n+1} = \frac{d^2}{dx^2} \sum_{n=1}^{\infty} \frac{x^{n+1}}{n+1}$$

Step 2: Determine the interval of convergence

Now, we will determine the interval of convergence of the rewritten power series. For a geometric series, we know that the series converges if \(|x| < 1\). So, our interval of convergence is \((-1, 1)\).

Step 3: Identify the known power series and find the function represented

We can recognize power series in second derivative as a geometric series without first term. The geometric series can be expressed as $$\sum_{n=0}^{\infty} x^n = \frac{1}{1-x}$$ Now, we will integrate twice to find the original series function, making sure to add the constants of integration: First integration: $$\int \frac{1}{1-x} \, dx = -\ln(1-x) + C_1$$ Second integration: $$\int -\ln(1-x) + C_1 \, dx = x\ln(1-x) - x + C_1x + C_2$$ Thus, the function represented by the given power series is: $$f(x) = x\ln(1-x) - x + C_1x + C_2$$ Within the interval of convergence, \((-1, 1)\), the given power series converges to the function \(f(x) = x\ln(1-x) - x + C_1x + C_2\).