Clarification on Limit Comparison Test

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For one of my classes we are using Manfred Stoll's, $\textit{Introduction to Real Analysis}$, and I had a question regarding the Limit Comparison Test. Here is the definition that they have:

Suppose $\sum a_{k}$ and $\sum b_{k}$ are two given series of positive real numbers.

\begin{align*} (a) & \: If \lim_{n \to \infty} \dfrac{a_{n}}{b_{n}}=L \: with \: 0<L<\infty, \: then \sum a_{k} \: converges \iff \sum b_{k} \: converges. \\ \\ (b) & \: If \lim_{n \to \infty} \dfrac{a_{n}}{b_{n}}=0 \: and \sum b_{k}, \: then \sum a_{k} \: converges \end{align*}

Okay. So the main problem that I have is that in various websites they have an additional condition (and if I recall right, in Calc III as well). This additional condition says that $\: If \lim_{n \to \infty} \dfrac{a_{n}}{b_{n}}=\infty \: and \sum b_{k} \: diverges, \: then \sum a_{k} \: diverges$.

So, I've been trying to rely on my textbook to prove a certain problem diverges. I have attempted to use the Dirichlet and Alternating Series Tests, but I don't have one of the necessary conditions to make them work. Thus, why is there a difference in this textbook and how some other writers/websites/etc., present the LCT? Any proof or clarification on this matter would be appreciated. I would rather rely on the textbook since it is rigorous, but at the same time it seems like my professor is okay with me using this modified version (even though it isn't given in the textbook).

!!!

*Edit* Additionally, does it have to do with that we are working with the finite? Since $\infty$ is an abstract/unattainable "number", is that why this book does not consider that as part of the test? On wikipedia it seems that they use the version in my textbook. However, on websites such as the one from Oregon State University and other websites they have the additional condition.

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i707107 gives the perfect answer! If $\lim a_n/b_n = \infty$ then given $\epsilon = 1, \exists N$ such that $a_n/b_n>1$ for all $n>N$ which imples $a_n > b_n$. Therefore if $\sum_n b_n$ diverges then so does $\sum_n a_n$.