Hi
How do I find limit of convegence of the series
a>1 ; n goes from 1 to infinity
sigma[n/(a^n)]
Thanks
$\displaystyle {{\varphi }_{n}}=\frac{n}{{{a}^{n}}}\implies \sqrt[n]{{{\varphi }_{n}}}=\frac{\sqrt[n]{n}}{a}\to \frac{1}{a}$ as $\displaystyle n\to\infty.$
Since $\displaystyle \frac{1}{a}<1,\,\forall \,a>1$ the series converges with given condition on $\displaystyle a.$
to evaluate the series, differentiate $\displaystyle 1+x+x^2 + \cdots = \frac{1}{1-x}, \ |x| < 1,$ to get $\displaystyle 1 + 2x + 3x^2 + \cdots = \frac{1}{(1-x)^2},$ now multiply by $\displaystyle x$ and then put $\displaystyle x=\frac{1}{a}$ to get $\displaystyle \sum_{n=1}^{\infty}\frac{n}{a^n}=\frac{a}{(a-1)^2}.$