Absolute convergence is important because it actually better captures the notion of "an infinite sum" in the informal sense than just convergence. For absolutely convergent series, you can talk about sums of n outputs of the sequence, for arbitrary natural n, and so, you can talk about the set of every such sum. You can then find the supremum of such a set, and the infimum of such a set. The two are equal if and only if the sequence of partial sums is absolutely convergent. The reason this captures the generalization better is because it is independent of the order of the elements, i.e it is permutation invariant, and it generalizes to uncountable sets better: rather than just sequences, you get a straightforward definition of integration, and other sum-like objects.
To proof minorant criterion. Can I said that as sum_bk diverge for all eps > 0 exist n0 in N such that for all n,m >= n0, sum_{k=m}^nb_k > eps so because a_k >= b_k for all k must be that sum_{k=m}^na_k >= sum_{k=m}^nb_k > eps so ak diverge?
0:30 absolute convergent
1:35 triangle inequality
2:15 counterexample immediately clear from Leibniz criterion
2:45 majorant criterion
4:20 proof
5:33 minorant criterion
Absolute convergence is important because it actually better captures the notion of "an infinite sum" in the informal sense than just convergence. For absolutely convergent series, you can talk about sums of n outputs of the sequence, for arbitrary natural n, and so, you can talk about the set of every such sum. You can then find the supremum of such a set, and the infimum of such a set. The two are equal if and only if the sequence of partial sums is absolutely convergent. The reason this captures the generalization better is because it is independent of the order of the elements, i.e it is permutation invariant, and it generalizes to uncountable sets better: rather than just sequences, you get a straightforward definition of integration, and other sum-like objects.
Perfect just finished watching Part 18.
My favorite mathematics channel.
I hope we can get a continuation for distribution theory.
Thank you very much! Distribution theory will be continued soon :)
Could you possibly make videos on numerical analysis and optimization methods? Thanks for the wonderful lessons!
Very good idea :)
Thank you
Great video, Cheers!!
To proof minorant criterion. Can I said that as sum_bk diverge for all eps > 0 exist n0 in N such that for all n,m >= n0, sum_{k=m}^nb_k > eps so because a_k >= b_k for all k must be that sum_{k=m}^na_k >= sum_{k=m}^nb_k > eps so ak diverge?