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Do (i), (ii) and (iii) apply to conditionally convergent series as well? I feel like they don't. But the book seems to say that they do because it doesn't "state otherwise".
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Why do you feel so, do you have a counterexample?I feel like they don't.
Is there any reordering involved in (i) to (iii)?
The first thing is not well-defined, and never appears in the problem.Yes there is. For (ii), it says ##u_1+u_2+...+v_1+v_2+...=u_1+v_1+u_2+v_2+...##.
While that is correct, it is way more complicated than necessary. Also, I would think most proofs of the squeeze theorem use one of those properties in some way, making the argument circular.If (ii) is true and if (i) is true for ##k=-1##, then (i) is true for all ##k##. Consider ##v_n=\frac{1}{2}u_n##, then using (ii), ##T=S-T##, so ##T=\frac{1}{2}S##. Similarly, we can prove (i) is true for all rational ##k##. Next, we use squeeze theorem to prove that (i) is true for all irrational ##k## too.
The first thing is not well-defined, and never appears in the problem.
Also, I would think most proofs of the squeeze theorem use one of those properties in some way, making the argument circular.
How does this step work? You would have to show that you can take out the factor (which is directly what you want to prove) or that limit and sum commute (which is not trivial, and in general false).Since (i) is true for rational ##k_1##,
##\lim_{k_1\rightarrow k}\Sigma k_1u_n=kS##. (And similarly true for ##k_2##.)
How does this step work? You would have to show that you can take out the factor (which is directly what you want to prove) or that limit and sum commute (which is not trivial, and in general false).
Anyway, you are making this way more complicated by going that route. You can directly prove it via the definition of a limit.
That part works, but how do you prove it is equal to ##\lim_{n \to \infty} \sum_{i=0}^n k u_i##?##\lim_{k_1\rightarrow k}\Sigma k_1u_n=\lim_{k_1\rightarrow k}k_1\Sigma u_n##, since (i) is true for rational ##k_1##
##=k\Sigma u_n=kS##
You can. For a finite sum, ##k \sum u_r = \sum k u_r## is trivial (distributive law), and you don't need it for "n=∞" because that never appears in the proof.I can't prove it directly.
That part works, but how do you prove it is equal to ##\lim_{n \to \infty} \sum_{i=0}^n k u_i##?
You can. For a finite sum, ##k \sum u_r = \sum k u_r## is trivial (distributive law), and you don't need it for "n=∞" because that never appears in the proof.
n is a natural number. Every natural number is finite.##\forall n>N## means we only need to consider finite values of ##n##?