Problem 59 - Calculus Early Transcendentals

Question

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

Step-by-Step Solution

Answer

Question: Identify the function represented by the following power series: $$\sum_{k=1}^{\infty} \frac{x^{k}}{k}$$ Answer: The function represented by the given power series is $$f(x) = C - \frac{1}{1-x}$$ where \(C\) is a constant.

Step 1: Identify and write down the geometric series formula

Recall the geometric series formula: $$\sum_{k=0}^{\infty} x^{k} = \frac{1}{1-x}, \quad |x|<1$$

Step 2: Derive the given power series

We will now derive the geometric series formula with respect to \(x\). Taking the derivative of \(\frac{1}{1-x}\) with respect to \(x\): $$\frac{d}{dx}\left(\frac{1}{1-x}\right) = \frac{1}{(1-x)^{2}}$$ Now, differentiating the power sum term-by-term and incrementing our starting index from \(k=0\) to \(k=1\) (noting that when \(k=0\), our term is simply a constant and will be eliminated when taking the derivative): $$\sum_{k=1}^{\infty} kx^{k-1} =\frac{1}{(1-x)^{2}}$$

Step 3: Integrate both sides with respect to x

Now, in order to identify the given power series, we need to integrate both sides of the equation with respect to \(x\). By doing so, we will get rid off the \(k\) in the numerator on the left side, and obtain the series given in the problem. $$\int\left(\sum_{k=1}^{\infty} kx^{k-1}\right)dx = \int\left(\frac{1}{(1-x)^{2}}\right)dx$$

Step 4: Perform the integration

Now, we will integrate the terms one by one. On the left side, integrate term-by-term: $$\sum_{k=1}^{\infty} \frac{x^{k}}{k} + C_1$$ On the right side: $$-\frac{1}{1-x} + C_2$$ where \(C_1\) and \(C_2\) are constants of integration.

Step 5: Identify the function represented by the power series

Combining both sides and equating the constants of integration, $$\sum_{k=1}^{\infty} \frac{x^{k}}{k} = C - \frac{1}{1-x}$$ where \(C = C_2 - C_1\). Thus, the function represented by the given power series is $$f(x) = C - \frac{1}{1-x}$$