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# 数学代写|实分析代写Real Analysis代考|The Difference

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## 数学代写|实分析代写Real Analysis代考|The Difference

As Riemann realized, the difference between the series in $(5.11)$ and in $(5.13)$ lies in the summation formed from just the positive (or negative) terms. In (5.13), the positive summands give a convergent series:
$$1+\frac{1}{4}+\frac{1}{16}+\frac{1}{64}+\cdots=\frac{4}{3} .$$
No matter how we rearrange our positive terms, they will never take us above $4 / 3$. The negative terms in this series will always subtract $2 / 3$. Any rearrangement will leave us with $4 / 3-2 / 3=2 / 3$
On the other hand, the series in (5.11) has positive terms whose sum diverges:
$$1+\frac{1}{5}+\frac{1}{9}+\frac{1}{13}+\cdots$$
The only thing that keeps the whole series from diverging is the presence of the negative terms that constantly compensate. The sum of the negative terms, taken on their own, must also diverge, otherwise they would not be sufficient to compensate for the diverging sum of positive terms. The difference between our series is that $(5.13)$ is absolutely convergent and (5.11) is not.

## 数学代写|实分析代写Real Analysis代考|Rearrangement with Conditional Convergence

If a series converges conditionally, then a rearrangement can change its value. How many possible values are there? Riemann realized that every real number can be obtained by rearranging such a series. You want to rearrange the series in (5.13) so that it converges to 1? We can do it. To 10.35 ? No problem. Sum it up to $-68+\sqrt{3}-e^\pi$ ? A piece of cake.
Theorem 5.3 (Riemann Rearrangement Theorem). If the series $a_1+a_2+a_3+\cdots$ converges conditionally, then for any real number $r$, we can find a rearrangement of this series that converges to $r$.

Rather than a formal proof, we shall see how this is done with an example. We shall take the series
$$1-\frac{1}{3}+\frac{1}{5}-\frac{1}{7}+\cdots$$
and rearrange it so that it converges to $\ln 2$ instead of $\pi / 4$. We separate the positive summands from the negative ones, keeping their relative order, and note that if we added up just the positive summands, the series would diverge. The same must be true for the negative summands. This is important.

## 数学代写|实分析代写Real Analysis代考|The Difference

$$1+\frac{1}{4}+\frac{1}{16}+\frac{1}{64}+\cdots=\frac{4}{3}$$

$$1+\frac{1}{5}+\frac{1}{9}+\frac{1}{13}+\cdots$$

## 数学代写|实分析代写Real Analysis代考|Rearrangement with Conditional Convergence

$$1-\frac{1}{3}+\frac{1}{5}-\frac{1}{7}+\cdots$$

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