Question #175376

Test the following series for convergence,

Σ n xn-1 , x > 0


1
Expert's answer
2021-03-30T07:28:03-0400

Let us test the series n=1nxn1, x>0,\sum_{n=1}^{\infty} n x^{n-1},\ x > 0, for convergence using D'Alembert criterion for positive series:


limn(n+1)xnnxn1=xlimnn+1n=x1=x\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<10<x<1 then the series is convergent. If x=1x=1 then the series n=1n\sum_{n=1}^{\infty} n is divergent because limnn=0.\lim\limits_{n\to\infty}n=\infty\ne 0.


Therefore, the series n=1nxn1, x>0,\sum_{n=1}^{\infty} n x^{n-1},\ x > 0, is convergent if and only if 0<x<10<x<1.


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