Series to integral conversion

  • Thread starter Damidami
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  • #1
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Hi!

Sorry if this is a bit trivial, I was wondering if there is a way of converting a series

\Sum_{n=1}^{+\infty} a_n
[TEX] \Sum_{n=1}^{+\infty} a_n [/TEX]

into an integral

\int_0^1 f(x) dx
[TEX]\int_0^1 f(x) dx[/TEX]

such that both are equal (give the same result). In that case, what is the relation between a_n and f(x) (are they some kind of reciproques?)
 

Answers and Replies

  • #2
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Yes there is. It's easier to do if you integrate over the interval [tex][0,\infty][/tex], but if you want to do it over [0,1] you can define [tex]f(x)=\sum_{ k = 1 }^\infty k ( k + 1 ) a_k \chi_{ ( \frac{1}{k+1}, \frac{1}{k} ] }[/tex], where [tex]\chi_A[/tex] is the characteristic function of the set A. Note that such a function is not unique, because the function [tex]g(x)=\sum_{k=1}^\infty k a_k \chi_{ (0,\frac{1}{k}] }[/tex] would also work.

You hit upon a somewhat deep result of analysis, which is that integration and summation are essentially the same thing. This is intuitively clear from the first time you learn about the Riemann integral as a limit of Riemann sums, but to really make it precise you need measure theory. If you're interested in this stuff, you should read about the Lebesgue integral. As it turns out, an infinite sum is just the Lebesgue integral over the natural numbers with respect to the counting measure. If you take this approach to summation, a whole slew of results such as Holder's and Minkowski's inequalities (of which the Cauchy-Schwartz inequality is a special case) and the criteria for interchanging double sums or limits with sums etc, follow quite simply as a result of the corresponding measure-theoretic facts.
 
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