Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

Think of the problem of convergence of the series $$ \sum_{k=1}^\infty (-1)^k a_k$$ Is it possible to consider the convergence of $\sum\limits_{k=1}^\infty (-1)^k b_k$ if $\lim \limits_{k\rightarrow\infty } \dfrac{a_k}{b_k}=$ limited and non-zero?

If not, is there any theorem that helps to facilitate the analysis of convergence when $a_k$ is a complicated function of $k$?

share|improve this question
1  
Are there any hypotheses at all for the $a_k$? Because otherwise, you're basically asking if there are theorems that help facilitate the analysis of convergence of any series. For without any hypotheses, the same theorems should also work for $a_k' = (-1)^k a_k$, and so you're considering $\sum a_k$. –  Najib Idrissi Jun 5 '13 at 9:12
add comment

2 Answers

up vote 2 down vote accepted

In the following I will assume that $a_n\ge0$; otherwise, as nick points out in a comment, you are asking about convergence criteria for arbitrary sequences.

The only general convergence criterion I know of is Leibiz' criterion.

There is no comparison theorem like the one you ask about. The series $$ \sum_{n=1}^\infty\frac{(-1)^n}{n}\quad\text{and}\quad\sum_{n=1}^\infty\frac{(-1)^n}{\sqrt n} $$ are both convergent by Leibniz criterion, but $$ \sum_{n=1}^\infty\frac{(-1)^n}{n-(-1)^n} $$ converges, while $$ \sum_{n=1}^\infty\frac{(-1)^n}{\sqrt n-(-1)^n} $$ diverges.

share|improve this answer
add comment

What about the Leibniz criterion http://en.wikipedia.org/wiki/Alternating_series_test

share|improve this answer
    
You can use [text](link) to obtain e.g. Leibniz criterion. Please edit your answer accordingly. –  Lord_Farin Jun 5 '13 at 8:59
add comment

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.