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How to characterize the speed of a divergent series ? I have a divergent series with a parameter $x$ in it. How can i characterize the speed of divergence for different $x$ ?

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While there are various rates of convergence, based on big-O, little-o, etc. notations, it will not make equal sense to talk about "speed of divergence". A divergent series need not grow uniformly. It might oscillate. Or there might be subsequences that grow and subsequences that oscillate in bounded fashion. One way to impose some consistency on the behavior of nonconvergent sequences is to look at its lim sup and lim inf. Perhaps you should explain the purpose of the characterization you seek. –  hardmath Jan 13 '11 at 4:57
@hardmath : the series is non-decreasing –  Rajesh D Jan 13 '11 at 5:22
This is a very vague question. Why don't you show us what series you have... –  Aryabhata Jan 13 '11 at 5:30
@Moron : I don't really have any series to work with, the only restriction is that it is non decreasing. –  Rajesh D Jan 13 '11 at 5:41

2 Answers 2

This is related to Hausdorff's "Pantachie" problem. Suppose $x_i$ is a monotone decreasing sequence whose sum is divergent. We say that a similar sequence $y_i$ diverges slower if $x_i/y_i \rightarrow \infty$. Example: $\sum n^{-1}$ diverges slower then $\sum n^{-0.5}$.

Similarly, if $x_i$ is a monotone decreasing sequence whose sum is convergent, a similar sequence $y_i$ converges more slowly if $y_i/x_i \rightarrow \infty$. Example: $\sum n^{-2}$ converges more slowly than $\sum n^{-3}$.

Hausdorff proved the following theorem: For any sequence of divergent (convergent) series, there's a sequence diverging (converging) slower than any of them. That means that there is no "expressible by finite strings" characterization of the speed of divergence (convergence), since such a characterization would not allow any series which is diverging (converging) slower.

For more on the subject, look up Hausdorff gaps.

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I guess ‘big O notation’ and its relatives are what you’re after, or something like that?

If $(a_n)$ is a sequence with $a_n \rightarrow \infty$ as $n \rightarrow \infty$, then this notation defines statements like $(a_n) = O(n^2)$, formalising the idea that “in the long run, the sequence $(a_n)$ grows no faster than the sequence $(n^2)$”.

(The use of ‘=’ in this notation is slightly confusing: the precise statement doesn’t assert that any two things are equal.)

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Under the given circumstances (the sequence $\{a_n\}$ tends monotonically to plus infinity), faster rate of divergence (growth) would be equivalent to faster convergence of $\{1/a_n\}$ to zero. –  hardmath Jan 13 '11 at 6:12

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