Answer to Question #175376 in Real Analysis for Anand

Question

Test the following series for convergence,

Σ n x^n-1 , x > 0

Answer 1

Let us test the series $\sum_{n=1}^{\infty} n x^{n-1},\ x > 0,$ for convergence using D'Alembert criterion for positive series:

$\lim\limits_{n\to\infty}\frac{(n+1)x^n}{nx^{n-1}}=x\lim\limits_{n\to\infty}\frac{n+1}{n}=x\cdot 1=x$

If $0<x<1$ then the series is convergent. If $x=1$ then the series $\sum_{n=1}^{\infty} n$ is divergent because $\lim\limits_{n\to\infty}n=\infty\ne 0.$

Therefore, the series $\sum_{n=1}^{\infty} n x^{n-1},\ x > 0,$ is convergent if and only if $0<x<1$.