Solution Found!
?(a) Suppose that \(\Sigma a_{n}\) and \(\Sigma b_{n}\) are series with positive terms
Chapter 10, Problem 49(choose chapter or problem)
(a) Suppose that \(\Sigma a_{n}\) and \(\Sigma b_{n}\) are series with positive terms and \(\Sigma b_{n}\) is divergent. Prove that if
\(\lim _{n \rightarrow \infty} \frac{a_{n}}{b_{n}}=\infty\)
then \(\Sigma a_{n}\) is also divergent.
(b) Use part (a) to show that the series diverges.
(i) \( \sum_{n=2}^{\infty} \frac{1}{\ln n}\)
(ii) \(\sum_{n=1}^{\infty} \frac{l n n}{n}\)
Questions & Answers
QUESTION:
(a) Suppose that \(\Sigma a_{n}\) and \(\Sigma b_{n}\) are series with positive terms and \(\Sigma b_{n}\) is divergent. Prove that if
\(\lim _{n \rightarrow \infty} \frac{a_{n}}{b_{n}}=\infty\)
then \(\Sigma a_{n}\) is also divergent.
(b) Use part (a) to show that the series diverges.
(i) \( \sum_{n=2}^{\infty} \frac{1}{\ln n}\)
(ii) \(\sum_{n=1}^{\infty} \frac{l n n}{n}\)
ANSWER:Step 1 of 3
a)
Given that and are series with positive terms and is divergent, and also we have:
Since, then for every, there is a, so that every we have:
From here, sincediverges, then so does the series.
Using the comparison test, we can say that diverges as well.
Thus, it is proved thatis also divergent.