Imum Coeli
Junior Member
- Joined
- Dec 3, 2012
- Messages
- 86
Hi I was wondering if someone could please check my proof?
Q: Let \(\displaystyle \sum_{n=1}^{\infty} a_n \) be an infinite series and let \(\displaystyle lim \:sup \; |a_n|^\frac{1}{n} = L \in [0,\infty] \) . Show the series convereges absolutely if \(\displaystyle L<1\)
A: Choose \(\displaystyle R\in \mathbb{R}\) with \(\displaystyle L<R<1\) then \(\displaystyle \exists\; N\in\mathbb{N} : sup\: \{|a_n|^\frac{1}{n} \: | \: n\geq k\} < R \) if \(\displaystyle k \geq N\)
So \(\displaystyle \exists \; n_1 \geq N \) such that \(\displaystyle |a_{n_1}|^\frac{1}{n_1} \leq R \implies |a_{n_1}| \leq R^{n_1}\)
So there exists a decreasing sequence \(\displaystyle n_1, n_2, n_3,...\) such that \(\displaystyle |a_{n_j}|^\frac{1}{n_j} \leq R \implies |a_{n_j}| \leq R^{n_j} \leq 1 \therefore (a_n)^\infty_{n=1} \) is bounded by 1 and hence converges absolutely, completeing the proof
Q: Let \(\displaystyle \sum_{n=1}^{\infty} a_n \) be an infinite series and let \(\displaystyle lim \:sup \; |a_n|^\frac{1}{n} = L \in [0,\infty] \) . Show the series convereges absolutely if \(\displaystyle L<1\)
A: Choose \(\displaystyle R\in \mathbb{R}\) with \(\displaystyle L<R<1\) then \(\displaystyle \exists\; N\in\mathbb{N} : sup\: \{|a_n|^\frac{1}{n} \: | \: n\geq k\} < R \) if \(\displaystyle k \geq N\)
So \(\displaystyle \exists \; n_1 \geq N \) such that \(\displaystyle |a_{n_1}|^\frac{1}{n_1} \leq R \implies |a_{n_1}| \leq R^{n_1}\)
So there exists a decreasing sequence \(\displaystyle n_1, n_2, n_3,...\) such that \(\displaystyle |a_{n_j}|^\frac{1}{n_j} \leq R \implies |a_{n_j}| \leq R^{n_j} \leq 1 \therefore (a_n)^\infty_{n=1} \) is bounded by 1 and hence converges absolutely, completeing the proof