Riemann series rearrangement theorem: Difference between revisions

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==Statement==
==Statement==


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Full timed transcript: <toggledisplay>
Full timed transcript: <toggledisplay>
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Vipul: Okay, so this talk is going to be about
Vipul: Okay, so this talk is going to be about
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Now the series is called *conditionally convergent*
Now the series is called *conditionally convergent*
if it


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with the degree difference test, which basically
with the degree difference test, which basically
again follows from the integral test. This
again follows from the integral test, this


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actual function summation does not converge.
absolute value summation does not converge.
You do have examples of series that are convergent
You do have examples of series that are convergent


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So a_1 + a_2 + ... + a_{n-1} and the next
So a_1 + a_2 + ... + a_{n-1} and the next
partial sum is a_1 + a_2 + ... + a_{n-1} plus
partial sum is a_1 + a_2 + ... + a_{n-1} + a_n.


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a_n. You've added a_n,right? If this partial
You've added a_n,right? If this partial
sum is in the ball, in this interval, and
sum is in the ball, in this interval, and


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not absolutely convergent. Similarly, negative
not absolutely convergent. Similarly, if the
terms
negative terms


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converged, the positive terms fall to converge
converged, the positive terms also converge
and then the absolute value will also have
and then the absolute value will also have


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it's just this. It is the limit as n approaches
it's just this. It is the limit as n approaches
inf. Inf just means... is the shorthand for
infinity. Inf just means... is the shorthand for


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Latest revision as of 00:53, 29 April 2014

ORIGINAL FULL PAGE: Riemann series rearrangement theorem
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This article describes a test that is used to determine, in some cases, whether a given infinite series or improper integral converges. It may help determine whether we have absolute convergence, conditional convergence, or neither.
View a complete list of convergence tests

Statement

Consider a series:

(Note: the starting point of the summation does not matter for the theorem).

Suppose the series is a conditionally convergent series: it is a convergent series but not an absolutely convergent series, i.e., the series does not converge.

Then, the following are true:

  1. The sub-series comprising only those s that are positive diverges.
  2. The sub-series comprising only those s that are negative diverges.
  3. Given any two elements in (i.e., they could be real numbers, or ) there exists a rearrangement of the s such that the limit inferior of the partial sums is and the limit superior of the partial sums is . In particular, since we are allowed to set , we can obtain a rearrangement that converges to any desired sum.
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Full timed transcript: [SHOW MORE]

Related facts

  • Levy-Steinitz theorem is a generalization to series of vectors in . The claim is that the set of possible sums of rearrangements of any series of vectors that are finite vectors, if non-empty, is an affine subspace of .

Proof

Proof of (1), (2), and (3)

(1) is true on account of convergence.

(2) and (3): We can show that if any one of the sums is finite, so is the other one, and both being finite would force absolute convergence.

Proof of (4)

We outline a constructive procedure to create the series. For simplicity, we will assume that none of the s are 0. The proof can be modified somewhat to include the case of some of the s being zero.

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Case of finite values of limit superior and limit inferior

Setup (unsorted version):

  • Pick the sub-series of all the positive value terms.
  • Pick the sub-series of all the negative value terms.

Setup (sorted version)

  • Arrange all the positive values among the s in decreasing order. We will always pick from the list in the order arranged, i.e., we will always pick the largest positive value currently in the list when asked to pick from the list. The fact that we can list all the positive s in decreasing order relies on .
  • Arrange all the negative values among the s in decreasing order of magnitude (hence, increasing order). We will always pick from the list in the order arranged, i.e., we will always pick the largest magnitude negative value currently in the list when asked to pick from the list. The fact that we can list all the negative s in increasing order relies on .

Rearrangement creation:

  • Begin by picking positive values (using the ordering of terms in the positive sub-series) and keep doing so until the partial sum so far is greater than . Note that we always reach such a point after picking finitely many positive values because the sum of all positive terms of the series is .
  • Now, start picking negative values (using the ordering of terms in the negative sub-series) till the overall partial sum (including the positive and negative values picked so far) is less than . Again, this is possible because the negative value terms add up to .
  • Now, resume picking positive values till the overall partial sum is greater than , then switch back to picking negative values, and so on.

If we are using the sorted version of the setup, we are always picking the largest magnitude term among those left within the terms of that sign. If we are using the unsorted version, we are picking the positive (respectively, negative) terms in the same order as in the original series. However, the nature of alternation between positive and negative terms may be quite far from the original series had.

Checking conditions:

  • Every term gets picked exactly once: We note that, because both the positive and negative term sums diverge, it is the case that we cycle back and forth infinitely many times. Each time we cycle, we pick at least one new term from each side, so it's clear that every term gets picked. The picking procedure makes sure every time is picked exactly once.
  • The limit superior is : We know that the partial sums become greater than or equal to infinitely often, so the limit superior is at least . The extent to which the partial sum can overshoot each time is bounded by the magnitude of the terms, which we have assumed approaches zero. This shows that the limit superior is exactly .
  • The limit inferior is : Similar reasoning as with the limit superior.

Case where we allow infinity for one or both the values

  • If the limit superior is and the limit inferior is finite: In this case, we construct an increasing sequence of finite numbers that approaches first (we can construct this sequence before looking at the s). Now, we modify the construction of the rearrangement series as follows: instead of trying to cross every time with the positive terms, we try to cross with the positive terms at the stage.
  • If the limit superior and limit inferior are both : We construct increasing sequences to in place of both and .
  • If the limit inferior is and the limit superior is finite: In this case, we construct a decreasing sequence of finite numbers that approaches first (we can construct this sequence without looking at the s). Now, we modify the construction of the rearrangement series as follows: instead of trying to cross every time with the negative terms, we try to cross with the negative terms at the stage.
  • If the limit superior and limit inferior are both : We construct decreasing sequences to in place of both and .
  • If the limit superior is and the limit inferior is : We construct an increasing sequence to in place of and a decreasing sequence to in place of .