The series you provided is:

$\sum_{n=1}^{\infty} n x^{n-1}$

Step 1: Recognizing the Series Type

This series is related to the geometric series, but with an additional factor of $n$ in each term. Let's break it down step by step.

The standard geometric series is:

$\sum_{n=0}^{\infty} x^n = \frac{1}{1-x}, \quad \text{for} \quad |x| < 1.$

If we differentiate both sides of this equation with respect to $x$, we get:

$\frac{d}{dx} \left( \sum_{n=0}^{\infty} x^n \right) = \frac{d}{dx} \left( \frac{1}{1-x} \right).$

The left-hand side gives:

$\sum_{n=1}^{\infty} n x^{n-1},$

and the right-hand side gives:

$\frac{1}{(1-x)^2}.$

Therefore, we have:

$\sum_{n=1}^{\infty} n x^{n-1} = \frac{1}{(1-x)^2}, \quad \text{for} \quad |x| < 1.$

Step 2: Radius of Convergence

The radius of convergence of the series can be determined from the fact that this is a differentiated geometric series. The geometric series converges for $|x| < 1$, and since differentiation does not affect the radius of convergence, the radius of convergence for this series is also $1$.

Thus, the radius of convergence is:

$R = 1.$

Step 3: Final Answer

• The sum of the series is:

$\sum_{n=1}^{\infty} n x^{n-1} = \frac{1}{(1-x)^2}, \quad \text{for} \quad |x| < 1.$

• The radius of convergence is:

$R = 1.$

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Would you like further details or clarification on any part of the solution?

Here are some related questions to explore:

1. What happens to the sum of this series when $x = 0$?
2. How would you find the sum of a series like $\sum_{n=0}^{\infty} n x^n$?
3. Can the geometric series $\sum_{n=0}^{\infty} x^n$ be used to solve other types of series?
4. How do you test the convergence of a series using the radius of convergence?
5. What is the behavior of the series as $x$ approaches 1?

Tip: For any power series, you can find the sum by differentiating a related geometric series or by recognizing it as a standard known series.