91Ó°ÊÓ

Show that a) if \(a_{n} \geq a_{n+1}>0\) for all \(n \in \mathbb{N}\) and the series \(\sum_{n=1}^{\infty} a_{n}\) converges, then \(a_{n}=\) \(o\left(\frac{1}{n}\right)\) as \(n \rightarrow \infty\) b) if \(b_{n}=o\left(\frac{1}{n}\right)\), one can always construct a convergent series \(\sum_{n=1}^{\infty} a_{n}\) such that \(b_{n}=o\left(a_{n}\right)\) as \(n \rightarrow \infty\) c) if a series \(\sum_{n=1}^{\infty} a_{n}\) with positive terms converges, then the series \(\sum_{n=1}^{\infty} A_{n}\) where \(A_{n}=\sqrt{\sum_{k=n}^{\infty} \alpha_{k}}-\sqrt{\sum_{k=n+1}^{\infty} a_{k}}\) also converges, and \(a_{n}=o\left(A_{n}\right)\) as \(n \rightarrow \infty\) d) if a series \(\sum_{n=1}^{\infty} a_{n}\) with positive terms diverges, then the series \(\sum_{n=2}^{\infty} A_{n}\) where \(A_{n}=\sqrt{\sum_{k=1}^{n} a_{k}}-\sqrt{\sum_{k=1}^{n-1} a_{k}}\) also diverges, and \(A_{n}=o\left(a_{n}\right)\) as \(n \rightarrow \infty\). It follows from c) and d) that no convergent (resp. divergent) series can serve as a universal standard of comparison to establish the convergence (resp. divergence) of other series.

Show that a) the series \(\sum_{n=1}^{\infty} \ln a_{n}\), where \(a_{n}>0, n \in \mathbb{N}\), converges if and only if the sequence \(\left\\{\Pi_{n}=a_{1} \cdots a_{n}\right\\}\) has a finite nonzero limit. b) the series \(\sum_{n=1}^{\infty} \ln \left(1+\alpha_{n}\right)\), where \(\left|\alpha_{n}\right|<1\), converges absolutely if and only if the series \(\sum_{n=1}^{\infty} \alpha_{n}\) converges absolutely. Hint: See part a) of Exercise \(5 .\)

An infinite product \(\prod_{k=1}^{\infty} e_{k}\) is said to converge if the sequence of numbers \(\Pi_{n}=\) \(\prod_{k=1}^{n} e_{k}\) has a finite nonzero limit \(\Pi .\) We then set \(\Pi=\prod_{k=1}^{\infty} e_{k}\). Show that a) if an infinite product \(\prod_{n=1}^{\infty} e_{n}\) converges, then \(e_{n} \Rightarrow 1\) as \(n \Rightarrow \infty\); b) if \(\forall n \in \mathbb{N}\left(e_{n}>0\right)\), then the infinite product \(\prod_{n=1}^{\infty} e_{n}\) converges if and only if the series \(\sum_{n=1}^{\infty} \ln e_{n}\) converges; c) if \(e_{n}=1+\alpha_{n}\) and the \(\alpha_{n}\) are all of the same sign, then the infinite product \(\prod_{n=1}^{\infty}\left(1+\alpha_{n}\right)\) converges if and only if the series \(\sum_{n=1}^{\infty} \alpha_{n}\) converges.