series convegnce

**omer.jack** · January 12th 2009, 09:38 AM

Hi

How do I find limit of convegence of the series

a>1 ; n goes from 1 to infinity

sigma[n/(a^n)]

Thanks

**Krizalid** · January 12th 2009, 09:43 AM

${{\varphi }_{n}}=\frac{n}{{{a}^{n}}}\implies \sqrt[n]{{{\varphi }_{n}}}=\frac{\sqrt[n]{n}}{a}\to \frac{1}{a}$ as $n\to\infty.$

Since $\frac{1}{a}<1,\,\forall \,a>1$ the series converges with given condition on $a.$

**NonCommAlg** · January 12th 2009, 10:11 AM

Originally Posted by omer.jack

Hi

How do I find limit of convegence of the series a>1 ; n goes from 1 to infinity sigma[n/(a^n)]

Thanks

to evaluate the series, differentiate $1+x+x^2 + \cdots = \frac{1}{1-x}, \ |x| < 1,$ to get $1 + 2x + 3x^2 + \cdots = \frac{1}{(1-x)^2},$ now multiply by $x$ and then put $x=\frac{1}{a}$ to get $\sum_{n=1}^{\infty}\frac{n}{a^n}=\frac{a}{(a-1)^2}.$

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