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Happiness
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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?Happiness said:I feel like they don't.
mfb said:Is there any reordering involved in (i) to (iii)?
The first thing is not well-defined, and never appears in the problem.Happiness said: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.
mfb said:The first thing is not well-defined, and never appears in the problem.
mfb said:There is still no rearrangement of the order of the elements in the series, e. g. u5 is always added after u4.
mfb said: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).Happiness said:Since (i) is true for rational ##k_1##,
##\lim_{k_1\rightarrow k}\Sigma k_1u_n=kS##. (And similarly true for ##k_2##.)
mfb said: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).
mfb said: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##?Happiness said:##\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.Happiness said:I can't prove it directly.
mfb said: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.Happiness said:##\forall n>N## means we only need to consider finite values of ##n##?
A conditionally convergent series is a mathematical series in which the sum of the terms converges to a finite value, but the series itself does not converge absolutely. This means that the sum of the absolute values of the terms diverges.
A series can be determined to be conditionally convergent by checking if the series converges when the absolute values of the terms are taken. If the series converges, it is absolutely convergent. If the series diverges, it is conditionally convergent.
Conditionally convergent series are significant because they do not follow the typical rules of convergence and can exhibit strange behavior. They also have important applications in mathematics, physics, and engineering.
Yes, a conditionally convergent series can be rearranged. However, the rearranged series may converge to a different value or may even diverge. This is known as the Riemann rearrangement theorem.
Conditionally convergent series are used in real-world applications to model various phenomena, such as alternating currents, financial markets, and signal processing. They are also used in numerical analysis and in the study of complex systems.