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 $x$ ?
what is the speed of a divergent series?
-
0@Moron : I don't really have any series to work with, the only restriction is that it is non decreasing. – 2011-01-13
2 Answers
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.
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.)
-
0Under 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. – 2011-01-13