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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 $k$?

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    $\begingroup$ 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 $\sum a_k$. $\endgroup$ – Najib Idrissi Jun 5 '13 at 9:12
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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.

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

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  • $\begingroup$ You can use [text](link) to obtain e.g. Leibniz criterion. Please edit your answer accordingly. $\endgroup$ – Lord_Farin Jun 5 '13 at 8:59

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