Convergence of Series

Real Analysis
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achilleas
Posts: 10
Joined: Tue Nov 10, 2015 5:35 pm

Convergence of Series

#1

Post by achilleas »

Let \(\alpha\) and \(\beta\) be real numbers, with \(\alpha>0\), and define the sequence \( \{a_n\}\) by \(a_1=\alpha\) and \(a_{n+1}=a_n\cdot \alpha^{a_n}\) for \(n=1,2,\ldots\). Study convergence of the series $$\sum_{n=1}^{\infty} \dfrac{a_n}{n^{\beta}}.$$
ZardoZ
Posts: 13
Joined: Wed Nov 11, 2015 12:47 pm

Re: Convergence of Series

#2

Post by ZardoZ »

We will first prove that \(a_n>0\) for all \(n\in\mathbb{N}\).

For \(n=1\), \(a_{1}=\alpha>0\), supposing that \(a_n>0\) for all \(n\in\mathbb{N}\), \(a_{n+1}=a_{n}{\alpha}^{a_{n}}>0\) which proves the induction.

If

\(\bullet\) \(\alpha=1\), then \(a_n=c\) for all \(n\in\mathbb{N}\) and some constant \(c\in\mathbb{R}\), so \(\displaystyle\sum_{n=1}^{\infty}\frac{a_{n}}{n^{\beta}}=c\sum_{n=1}^{\infty}\frac{1}{n^{\beta}}\) which is convergent for \(\beta>1\).

\(\bullet\) \(\alpha\in(0,1)\), then the sequence \(a_{n}\) is stricly decreasing since \(\displaystyle \log\left(\frac{a_{n+1}}{a_{n}}\right)=a_{n}\cdot \log(\alpha)<0\Rightarrow a_{n+1}<a_{n}\) and by the D'Alembert convergence criterion we get

\[\overline{\lim_{n\to\infty}}\left|\frac{\frac{a_{n+1}}{(n+1)^{\beta}}}{\frac{a_{n}}{n^{\beta}}}\right|=\overline{\lim_{n\to\infty}}\left|\frac{a_{n+1}}{a_{n}}\left(1+\frac{1}{n}\right)^{\beta}\right|<1\]

and so the series is convergent for all \(\beta\).

\(\bullet\) \(\alpha>1\), then \(a_n\) is strictly increasing since \(\displaystyle \log\left(\frac{a_{n+1}}{a_{n}}\right)=a_{n}\cdot \log(\alpha)>0\Rightarrow a_{n+1}>a_{n}\) and by the D'Alembert convergence criterion we get

\[\overline{\lim_{n\to\infty}}\left|\frac{\frac{a_{n+1}}{(n+1)^{\beta}}}{\frac{a_{n}}{n^{\beta}}}\right|=\overline{\lim_{n\to\infty}}\left|\frac{a_{n+1}}{a_{n}}\left(1+\frac{1}{n}\right)^{\beta}\right|>1\]

and so the series is divergent for all \(\beta\).
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