Talk:Summation by parts
This article has not yet been rated on Wikipedia's content assessment scale. |
First, there is a typo in the formula for the partial sums, in the last formula of the "Methods" section, the first b_N should be a B_N. Next, I believe that the hypotheses in the "Applications" section is not sufficient for the result stated. In the current case, you can take the sequence a_n to be a constant sequence, say identically 1, and the sequence b_n to be the following: 1, -1/2, -1/2, 1/3, 1/3, 1/3, ... (there are n terms each of 1/n, with alternating signs between each group). The sequence a_nb_n approaches 0, the series (a_(n+1)-a_n) is absolutely convergent, and the partial sums B_n are uniformly bounded (they are all between 0 and 1). However, the sum a_nb_n does not converge, since it is just the sum of b_n which oscillates infinitely often between 0 and 1. The correct hypothesis for convergence should be to replace "a_nb_n approches 0", by the condition that "a_n approaches 0". This can be seen by writing the Cauchy sequence of the partial sums in the summation by parts formulation and then estimating each part.
I think there's some typo in the second equation, introducing the \Delta operator. The first term on the right hand side, f_{k+1} g_{k+1} should be f_{n+1} g_{n+1} - f_m g_m. The summation variable k is not bound... / nisse@lysator.liu.se
formula for shifted index
editThe formula is listed for this sum:
but I think it'd be nice to also list the formula for this sum:
which is slightly different, but in the same general form. Lavaka (talk) 18:11, 15 January 2008 (UTC)
which results from iterated application of the initial formula.
"The auxiliary quantities are Newton series:" Here, the link to the newton series is actually a link to the difference operator. I think it should be fixed. I don't know how though..
Applications
editAgreed, as it is stated now, the hypotheses are incorrect. The reason the proof seems to work is that there was an accidental change from using to which makes it seem as though the proof is correct.
infinite series; conditional convergence
editDoes this property hold for infinite series in general? I do know that some of the basic algebraic properties - associativity, commutativity - that goes into the proof of this property, do not apply to conditionally convergent series. Just wondering if this article is complete, or if it needs additional conditioning on its applicability. 134.204.1.226 (talk) 16:29, 25 May 2023 (UTC)