Given the geometric series $\sum n_0=\infty x^n$, find the sum of the series $\sum n_1=\infty n{x}^{n-1}$, where $\left|x\right|<1$.

• A. $\frac{x}{1-x}$
• B. $\frac{1}{1-x}$
• C. $\frac{x}{{1-x}^2}$
• D. $\frac{1}{{1-x}^2}$

Answer: (D)

Verified step by step guidance

1. Step 1: Recall the formula for the sum of a geometric series. For the series \( \sum_{n=0}^{\infty} x^n \), the sum is \( \frac{1}{1-x} \) when \( |x| < 1 \). This will be useful for manipulating the given series.
2. Step 2: Analyze the series \( \sum_{n=1}^{\infty} n x^{n-1} \). Notice that the term \( n x^{n-1} \) involves the derivative of the geometric series \( \sum_{n=0}^{\infty} x^n \). Specifically, differentiate \( \frac{1}{1-x} \) with respect to \( x \).
3. Step 3: Compute the derivative of \( \frac{1}{1-x} \). Using the chain rule, the derivative is \( \frac{d}{dx} \left( \frac{1}{1-x} \right) = \frac{1}{(1-x)^2} \). This derivative represents \( \sum_{n=1}^{\infty} n x^{n-1} \).
4. Step 4: Substitute the result of the derivative into the series. The sum \( \sum_{n=1}^{\infty} n x^{n-1} \) is equal to \( \frac{1}{(1-x)^2} \).
5. Step 5: Verify the conditions for convergence. Since \( |x| < 1 \), the series converges, and the final expression \( \frac{1}{(1-x)^2} \) is valid.