Assume f(x) positive and continuous,for improper integrals of the first kind, is there an analog of the n-th term divergence test for series ? It would say:
Give a proof or counterexample.
I think ,but I can not give a counterexample.
Assume f(x) positive and continuous,for improper integrals of the first kind, is there an analog of the n-th term divergence test for series ? It would say:
Give a proof or counterexample.
I think ,but I can not give a counterexample.
Yes, that's correct. If converges, and exists, then it must be zero. Otherwise, stays above some positive value when x is large enough, say , and the integral is greater than the divergent integral .
The assumption that exists is critical. There are functions whose integral converges but have no limit as x goes to infinity. As a hint, consider the function defined on [0,1], when and when . You can pick any number between 0 and 1 to be the value of its integral on [0,1].
- Hollywood
No, he didn't say, but it's the additional assumption he needs to prove his result.
Without that additional assumption, the result is false. And we know a little bit about what a counterexample might look like (or what it doesn't look like).
I guess my post was misleading - sorry about that.
- Hollywood
Yes thanks for the correction.
A simpler example can be a function which is zero everywhere except in [n, n+1/n^3], and inside such intervals it forms a triangle who's peak is at f(x) = 1. Basically, it would be a series of thinner and thinner triangles, each of the same height yet increasingly skinnier enough for convergence.
It sounds like he wants something positive. You both have the right idea, though - take a function whose integral converges and add spikes. The spikes should become narrower so that the integral of the spikes converges, but the height of the spikes can not go to zero, meaning the limit of the function doesn't exist.
- Hollywood