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Question about series rearrangement in Baby Rudin (theorem 3.54).
$begingroup$
I have trouble following the following theorem in Rudin:
Let $sum a_n$ be a series of real numbers which converges, but not absolutely. Suppose
$$-infty leq alpha leq beta leq +infty.$$
Then there exists a rearrangement $sum a_n^prime$ with partial sums $s_n^prime$ such that
$$lim_{ntoinfty}inf s_n^prime = alpha, mbox{ and } lim_{ntoinfty}sup s_n^prime = beta. tag{24}$$
Here's the proof:
Let $$p_n = frac{|a_n| + a_n}{2}, q_n = frac{|a_n| - a_n}{2} (n = 1, 2, 3, ldots). $$
Then $p_n - q_n = a_n$, $p_n + q_n = |a_n|$, $p_n geq 0$, $q_n geq 0$. The series $sum p_n$, $sum q_n$ must both diverge.
For if both were convergent, then $$sum left( p_n + q_n right) = sum |a_n|$$ would converge, contrary to hypothesis. Since $$ sum_{n=1}^N a_n = sum_{n=1}^N left( p_n - q_n right) = sum_{n=1}^N p_n - sum_{n=1}^N q_n,$$ divergence of $sum p_n$ and convergence of $sum q_n$ (or vice versa) implies divergence of $sum a_n$, again contrary to hypothesis.
Now let $P_1, P_2, P_3, ldots$ denote the non-negative terms of $sum a_n$, in the order in which they occur, and let $Q_1, Q_2, Q_3, ldots$ be the absolute values of the negative terms of $sum a_n$, also in their original order.
The series $sum P_n$, $sum Q_n$ differ from $sum p_n$, $sum q_n$ only by zero terms, and are therefore divergent.
We shall construct sequences ${m_n }$, ${k_n}$, such that the series $$ P_1 + cdots + P_{m_1} - Q_1 - cdots - Q_{k_1} + P_{m_1 + 1} + cdots + P_{m_2} - Q_{k_1 + 1} - cdots - Q_{k_2} + cdots tag{25}, $$ which clearly is a rearrangement of $sum a_n$, satisfies (24).
Choose real-valued sequences ${ alpha_n }$, ${ beta_n }$ such that $alpha_n rightarrow alpha$, $beta_n rightarrow beta$, $alpha_n < beta_n$, $beta_1 > 0$.
Let $m_1$, $k_1$ be the smallest integers such that $$P_1 + cdots + P_{m_1} > beta_1,$$ $$P_1 + cdots + P_{m_1} - Q_1 - cdots - Q_{k_1} < alpha_1;$$ let $m_2$, $k_2$ be the smallest integers such that $$P_1 + cdots + P_{m_1} - Q_1 - cdots - Q_{k_1} + P_{m_1 + 1} + cdots + P_{m_2} > beta_2,$$ $$P_1 + cdots + P_{m_1} - Q_1 - cdots - Q_{k_1} + P_{m_1 + 1} + cdots + P_{m_2} - Q_{k_1 + 1} - cdots - Q_{k_2} < alpha_2;$$ and continue in this way. This is possible since $sum P_n$, $sum Q_n$ diverge.
If $x_n$, $y_n$ denote the partial sums of (25) whose last terms are $P_{m_n}$, $-Q_{k_n}$, then $$ | x_n - beta_n | leq P_{m_n}, |y_n - alpha_n | leq Q_{k_n}. $$ Since $P_n rightarrow 0$, $Q_n rightarrow 0$ as $n rightarrow infty$, we see that $x_n rightarrow beta$, $y_n rightarrow alpha$.
Finally, it is clear that no number less than $alpha$ or greater than $beta$ can be a subsequential limit of the partial sums of (25).
I don't understand the last two lines of the proof (put in bold). I'm aware that this question was already asked on this forums, but I didn't understand the answers provided in this question, so that's why I'm writing my own question.
Thanks in advance.
real-analysis sequences-and-series proof-explanation
edited Jul 24 '18 at 18:40
zipirovich
11.3k11731
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add a comment |
$begingroup$
I have trouble following the following theorem in Rudin:
Let $sum a_n$ be a series of real numbers which converges, but not absolutely. Suppose
$$-infty leq alpha leq beta leq +infty.$$
Then there exists a rearrangement $sum a_n^prime$ with partial sums $s_n^prime$ such that
$$lim_{ntoinfty}inf s_n^prime = alpha, mbox{ and } lim_{ntoinfty}sup s_n^prime = beta. tag{24}$$
Here's the proof:
Let $$p_n = frac{|a_n| + a_n}{2}, q_n = frac{|a_n| - a_n}{2} (n = 1, 2, 3, ldots). $$
Then $p_n - q_n = a_n$, $p_n + q_n = |a_n|$, $p_n geq 0$, $q_n geq 0$. The series $sum p_n$, $sum q_n$ must both diverge.
For if both were convergent, then $$sum left( p_n + q_n right) = sum |a_n|$$ would converge, contrary to hypothesis. Since $$ sum_{n=1}^N a_n = sum_{n=1}^N left( p_n - q_n right) = sum_{n=1}^N p_n - sum_{n=1}^N q_n,$$ divergence of $sum p_n$ and convergence of $sum q_n$ (or vice versa) implies divergence of $sum a_n$, again contrary to hypothesis.
Now let $P_1, P_2, P_3, ldots$ denote the non-negative terms of $sum a_n$, in the order in which they occur, and let $Q_1, Q_2, Q_3, ldots$ be the absolute values of the negative terms of $sum a_n$, also in their original order.
The series $sum P_n$, $sum Q_n$ differ from $sum p_n$, $sum q_n$ only by zero terms, and are therefore divergent.
We shall construct sequences ${m_n }$, ${k_n}$, such that the series $$ P_1 + cdots + P_{m_1} - Q_1 - cdots - Q_{k_1} + P_{m_1 + 1} + cdots + P_{m_2} - Q_{k_1 + 1} - cdots - Q_{k_2} + cdots tag{25}, $$ which clearly is a rearrangement of $sum a_n$, satisfies (24).
Choose real-valued sequences ${ alpha_n }$, ${ beta_n }$ such that $alpha_n rightarrow alpha$, $beta_n rightarrow beta$, $alpha_n < beta_n$, $beta_1 > 0$.
Let $m_1$, $k_1$ be the smallest integers such that $$P_1 + cdots + P_{m_1} > beta_1,$$ $$P_1 + cdots + P_{m_1} - Q_1 - cdots - Q_{k_1} < alpha_1;$$ let $m_2$, $k_2$ be the smallest integers such that $$P_1 + cdots + P_{m_1} - Q_1 - cdots - Q_{k_1} + P_{m_1 + 1} + cdots + P_{m_2} > beta_2,$$ $$P_1 + cdots + P_{m_1} - Q_1 - cdots - Q_{k_1} + P_{m_1 + 1} + cdots + P_{m_2} - Q_{k_1 + 1} - cdots - Q_{k_2} < alpha_2;$$ and continue in this way. This is possible since $sum P_n$, $sum Q_n$ diverge.
If $x_n$, $y_n$ denote the partial sums of (25) whose last terms are $P_{m_n}$, $-Q_{k_n}$, then $$ | x_n - beta_n | leq P_{m_n}, |y_n - alpha_n | leq Q_{k_n}. $$ Since $P_n rightarrow 0$, $Q_n rightarrow 0$ as $n rightarrow infty$, we see that $x_n rightarrow beta$, $y_n rightarrow alpha$.
Finally, it is clear that no number less than $alpha$ or greater than $beta$ can be a subsequential limit of the partial sums of (25).
I don't understand the last two lines of the proof (put in bold). I'm aware that this question was already asked on this forums, but I didn't understand the answers provided in this question, so that's why I'm writing my own question.
Thanks in advance.
real-analysis sequences-and-series proof-explanation
edited Jul 24 '18 at 18:40
zipirovich
11.3k11731
$endgroup$
$begingroup$
I'm not sure what you want to number or I'd edit the question myself. You can number an equation as (24) by adding tag{24} at the end of it.
$endgroup$
– saulspatz
Jul 24 '18 at 15:29
$begingroup$
Thanks for your comment. I edited the post.
$endgroup$
– user370967
Jul 24 '18 at 15:34
add a comment |
$begingroup$
I have trouble following the following theorem in Rudin:
Let $sum a_n$ be a series of real numbers which converges, but not absolutely. Suppose
$$-infty leq alpha leq beta leq +infty.$$
Then there exists a rearrangement $sum a_n^prime$ with partial sums $s_n^prime$ such that
$$lim_{ntoinfty}inf s_n^prime = alpha, mbox{ and } lim_{ntoinfty}sup s_n^prime = beta. tag{24}$$
Here's the proof:
Let $$p_n = frac{|a_n| + a_n}{2}, q_n = frac{|a_n| - a_n}{2} (n = 1, 2, 3, ldots). $$
Then $p_n - q_n = a_n$, $p_n + q_n = |a_n|$, $p_n geq 0$, $q_n geq 0$. The series $sum p_n$, $sum q_n$ must both diverge.
For if both were convergent, then $$sum left( p_n + q_n right) = sum |a_n|$$ would converge, contrary to hypothesis. Since $$ sum_{n=1}^N a_n = sum_{n=1}^N left( p_n - q_n right) = sum_{n=1}^N p_n - sum_{n=1}^N q_n,$$ divergence of $sum p_n$ and convergence of $sum q_n$ (or vice versa) implies divergence of $sum a_n$, again contrary to hypothesis.
Now let $P_1, P_2, P_3, ldots$ denote the non-negative terms of $sum a_n$, in the order in which they occur, and let $Q_1, Q_2, Q_3, ldots$ be the absolute values of the negative terms of $sum a_n$, also in their original order.
The series $sum P_n$, $sum Q_n$ differ from $sum p_n$, $sum q_n$ only by zero terms, and are therefore divergent.
We shall construct sequences ${m_n }$, ${k_n}$, such that the series $$ P_1 + cdots + P_{m_1} - Q_1 - cdots - Q_{k_1} + P_{m_1 + 1} + cdots + P_{m_2} - Q_{k_1 + 1} - cdots - Q_{k_2} + cdots tag{25}, $$ which clearly is a rearrangement of $sum a_n$, satisfies (24).
Choose real-valued sequences ${ alpha_n }$, ${ beta_n }$ such that $alpha_n rightarrow alpha$, $beta_n rightarrow beta$, $alpha_n < beta_n$, $beta_1 > 0$.
Let $m_1$, $k_1$ be the smallest integers such that $$P_1 + cdots + P_{m_1} > beta_1,$$ $$P_1 + cdots + P_{m_1} - Q_1 - cdots - Q_{k_1} < alpha_1;$$ let $m_2$, $k_2$ be the smallest integers such that $$P_1 + cdots + P_{m_1} - Q_1 - cdots - Q_{k_1} + P_{m_1 + 1} + cdots + P_{m_2} > beta_2,$$ $$P_1 + cdots + P_{m_1} - Q_1 - cdots - Q_{k_1} + P_{m_1 + 1} + cdots + P_{m_2} - Q_{k_1 + 1} - cdots - Q_{k_2} < alpha_2;$$ and continue in this way. This is possible since $sum P_n$, $sum Q_n$ diverge.
If $x_n$, $y_n$ denote the partial sums of (25) whose last terms are $P_{m_n}$, $-Q_{k_n}$, then $$ | x_n - beta_n | leq P_{m_n}, |y_n - alpha_n | leq Q_{k_n}. $$ Since $P_n rightarrow 0$, $Q_n rightarrow 0$ as $n rightarrow infty$, we see that $x_n rightarrow beta$, $y_n rightarrow alpha$.
Finally, it is clear that no number less than $alpha$ or greater than $beta$ can be a subsequential limit of the partial sums of (25).
I don't understand the last two lines of the proof (put in bold). I'm aware that this question was already asked on this forums, but I didn't understand the answers provided in this question, so that's why I'm writing my own question.
Thanks in advance.
real-analysis sequences-and-series proof-explanation
edited Jul 24 '18 at 18:40
zipirovich
11.3k11731
$endgroup$
I have trouble following the following theorem in Rudin:
Let $sum a_n$ be a series of real numbers which converges, but not absolutely. Suppose
$$-infty leq alpha leq beta leq +infty.$$
Then there exists a rearrangement $sum a_n^prime$ with partial sums $s_n^prime$ such that
$$lim_{ntoinfty}inf s_n^prime = alpha, mbox{ and } lim_{ntoinfty}sup s_n^prime = beta. tag{24}$$
Here's the proof:
Let $$p_n = frac{|a_n| + a_n}{2}, q_n = frac{|a_n| - a_n}{2} (n = 1, 2, 3, ldots). $$
Then $p_n - q_n = a_n$, $p_n + q_n = |a_n|$, $p_n geq 0$, $q_n geq 0$. The series $sum p_n$, $sum q_n$ must both diverge.
For if both were convergent, then $$sum left( p_n + q_n right) = sum |a_n|$$ would converge, contrary to hypothesis. Since $$ sum_{n=1}^N a_n = sum_{n=1}^N left( p_n - q_n right) = sum_{n=1}^N p_n - sum_{n=1}^N q_n,$$ divergence of $sum p_n$ and convergence of $sum q_n$ (or vice versa) implies divergence of $sum a_n$, again contrary to hypothesis.
Now let $P_1, P_2, P_3, ldots$ denote the non-negative terms of $sum a_n$, in the order in which they occur, and let $Q_1, Q_2, Q_3, ldots$ be the absolute values of the negative terms of $sum a_n$, also in their original order.
The series $sum P_n$, $sum Q_n$ differ from $sum p_n$, $sum q_n$ only by zero terms, and are therefore divergent.
We shall construct sequences ${m_n }$, ${k_n}$, such that the series $$ P_1 + cdots + P_{m_1} - Q_1 - cdots - Q_{k_1} + P_{m_1 + 1} + cdots + P_{m_2} - Q_{k_1 + 1} - cdots - Q_{k_2} + cdots tag{25}, $$ which clearly is a rearrangement of $sum a_n$, satisfies (24).
Choose real-valued sequences ${ alpha_n }$, ${ beta_n }$ such that $alpha_n rightarrow alpha$, $beta_n rightarrow beta$, $alpha_n < beta_n$, $beta_1 > 0$.
Let $m_1$, $k_1$ be the smallest integers such that $$P_1 + cdots + P_{m_1} > beta_1,$$ $$P_1 + cdots + P_{m_1} - Q_1 - cdots - Q_{k_1} < alpha_1;$$ let $m_2$, $k_2$ be the smallest integers such that $$P_1 + cdots + P_{m_1} - Q_1 - cdots - Q_{k_1} + P_{m_1 + 1} + cdots + P_{m_2} > beta_2,$$ $$P_1 + cdots + P_{m_1} - Q_1 - cdots - Q_{k_1} + P_{m_1 + 1} + cdots + P_{m_2} - Q_{k_1 + 1} - cdots - Q_{k_2} < alpha_2;$$ and continue in this way. This is possible since $sum P_n$, $sum Q_n$ diverge.
If $x_n$, $y_n$ denote the partial sums of (25) whose last terms are $P_{m_n}$, $-Q_{k_n}$, then $$ | x_n - beta_n | leq P_{m_n}, |y_n - alpha_n | leq Q_{k_n}. $$ Since $P_n rightarrow 0$, $Q_n rightarrow 0$ as $n rightarrow infty$, we see that $x_n rightarrow beta$, $y_n rightarrow alpha$.
Finally, it is clear that no number less than $alpha$ or greater than $beta$ can be a subsequential limit of the partial sums of (25).
I don't understand the last two lines of the proof (put in bold). I'm aware that this question was already asked on this forums, but I didn't understand the answers provided in this question, so that's why I'm writing my own question.
Thanks in advance.
real-analysis sequences-and-series proof-explanation
real-analysis sequences-and-series proof-explanation
edited Jul 24 '18 at 18:40
zipirovich
11.3k11731
edited Jul 24 '18 at 18:40
zipirovich
11.3k11731
edited Jul 24 '18 at 18:40
zipirovich
11.3k11731
edited Jul 24 '18 at 18:40
zipirovich
11.3k11731
edited Jul 24 '18 at 18:40
zipirovich
11.3k11731
11.3k11731
asked Jul 24 '18 at 15:25
user370967
$begingroup$
I'm not sure what you want to number or I'd edit the question myself. You can number an equation as (24) by adding tag{24} at the end of it.
$endgroup$
– saulspatz
Jul 24 '18 at 15:29
$begingroup$
Thanks for your comment. I edited the post.
$endgroup$
– user370967
Jul 24 '18 at 15:34
add a comment |
$begingroup$
I'm not sure what you want to number or I'd edit the question myself. You can number an equation as (24) by adding tag{24} at the end of it.
$endgroup$
– saulspatz
Jul 24 '18 at 15:29
$begingroup$
Thanks for your comment. I edited the post.
$endgroup$
– user370967
Jul 24 '18 at 15:34
$begingroup$
I'm not sure what you want to number or I'd edit the question myself. You can number an equation as (24) by adding tag{24} at the end of it.
$endgroup$
– saulspatz
Jul 24 '18 at 15:29
$begingroup$
I'm not sure what you want to number or I'd edit the question myself. You can number an equation as (24) by adding tag{24} at the end of it.
$endgroup$
– saulspatz
Jul 24 '18 at 15:29
$begingroup$
Thanks for your comment. I edited the post.
$endgroup$
– user370967
Jul 24 '18 at 15:34
$begingroup$
Thanks for your comment. I edited the post.
$endgroup$
– user370967
Jul 24 '18 at 15:34
add a comment |
2 Answers
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First, let me make sure it's clear what's happening behind all those formulas. The rearrangement that allegedly works goes like this: First take just enough positive terms from your given series to produce a partial sum $>beta$. (You can do that because the series of all the positive terms diverges.) After that, put just enough negative terms to bring the partial sum down below $alpha$ (possible because the series of all the negative terms diverges). Then resume putting just enough positive terms to bring the partial sum back up above $beta$. Continue working back and forth like this.
Notice that I said just enough terms at every stage. That ensures that, when you get a partial sum $s$ above $beta$, it won't be too far above $beta$; the difference $s-beta$ will be at most the last term you added, because otherwise you could have stopped adding positive terms sooner. Similarly, when the partial sum goes below $alpha$, the difference will be (in absolute value) at most the (absolute value of the) last term you added.
But your original series converged (conditionally), so the terms approach zero. That means that the amounts by which you overshoot $beta$ and undershoot $alpha$ are eventually arbitrarily small as you perform more and more stages of the process. And that's what those last two lines in Rudin's proof are saying is "clear".
answered Jul 28 '18 at 0:59
Andreas BlassAndreas Blass
50.5k452109
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I will consider the case $beta in mathbb{R}$ here.
Rudin showed that $x_n to beta$ on p.77.
Let $epsilon$ be an arbitrary positive real number.
Then, there exists a natural number $N$ such that $$n geq N implies beta - epsilon < x_n < beta + epsilon.$$
By the construction of ${s'_n}$, the follwoing inequalities hold:
$$beta + epsilon > x_N = s'_{m_1 + k_1 + cdots + m_{N-1} + k_{N-1} + m_N} > s'_{m_1 + k_1 + cdots + m_{N-1} + k_{N-1} + m_N + 1} \> cdots > s'_{m_1 + k_1 + cdots + m_{N-1} + k_{N-1} + m_N + k_N} = y_N < s'_{m_1 + k_1 + cdots + m_{N-1} + k_{N-1} + m_N + k_N+1} < cdots < s'_{m_1 + k_1 + cdots + m_{N-1} + k_{N-1} + m_N + k_N+(m_{N+1}-1)} < s'_{m_1 + k_1 + cdots + m_{N-1} + k_{N-1} + m_N + k_N+m_{N+1}} = x_{N+1} < beta + epsilon.$$
Let $M := m_1 + k_1 + cdots + m_{N-1} + k_{N-1} + m_N$.
The above inequalities say that $s'_n < beta + epsilon$ for all $n in {M, M+1, cdots, M + k_N + m_{N+1}}$.
And we see that the following ineqality holds:
$s'_n < beta + epsilon$ for all $n in {M, M+1, cdots}$.
answered Jan 30 at 9:28
tchappy hatchappy ha
778412
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Thank you for your answer. I will look at it once I find time :)
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– user370967
Jan 30 at 17:52
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2 Answers
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2 Answers
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$begingroup$
First, let me make sure it's clear what's happening behind all those formulas. The rearrangement that allegedly works goes like this: First take just enough positive terms from your given series to produce a partial sum $>beta$. (You can do that because the series of all the positive terms diverges.) After that, put just enough negative terms to bring the partial sum down below $alpha$ (possible because the series of all the negative terms diverges). Then resume putting just enough positive terms to bring the partial sum back up above $beta$. Continue working back and forth like this.
Notice that I said just enough terms at every stage. That ensures that, when you get a partial sum $s$ above $beta$, it won't be too far above $beta$; the difference $s-beta$ will be at most the last term you added, because otherwise you could have stopped adding positive terms sooner. Similarly, when the partial sum goes below $alpha$, the difference will be (in absolute value) at most the (absolute value of the) last term you added.
But your original series converged (conditionally), so the terms approach zero. That means that the amounts by which you overshoot $beta$ and undershoot $alpha$ are eventually arbitrarily small as you perform more and more stages of the process. And that's what those last two lines in Rudin's proof are saying is "clear".
answered Jul 28 '18 at 0:59
Andreas BlassAndreas Blass
50.5k452109
$endgroup$
add a comment |
$begingroup$
First, let me make sure it's clear what's happening behind all those formulas. The rearrangement that allegedly works goes like this: First take just enough positive terms from your given series to produce a partial sum $>beta$. (You can do that because the series of all the positive terms diverges.) After that, put just enough negative terms to bring the partial sum down below $alpha$ (possible because the series of all the negative terms diverges). Then resume putting just enough positive terms to bring the partial sum back up above $beta$. Continue working back and forth like this.
Notice that I said just enough terms at every stage. That ensures that, when you get a partial sum $s$ above $beta$, it won't be too far above $beta$; the difference $s-beta$ will be at most the last term you added, because otherwise you could have stopped adding positive terms sooner. Similarly, when the partial sum goes below $alpha$, the difference will be (in absolute value) at most the (absolute value of the) last term you added.
But your original series converged (conditionally), so the terms approach zero. That means that the amounts by which you overshoot $beta$ and undershoot $alpha$ are eventually arbitrarily small as you perform more and more stages of the process. And that's what those last two lines in Rudin's proof are saying is "clear".
answered Jul 28 '18 at 0:59
Andreas BlassAndreas Blass
50.5k452109
$endgroup$
add a comment |
$begingroup$
First, let me make sure it's clear what's happening behind all those formulas. The rearrangement that allegedly works goes like this: First take just enough positive terms from your given series to produce a partial sum $>beta$. (You can do that because the series of all the positive terms diverges.) After that, put just enough negative terms to bring the partial sum down below $alpha$ (possible because the series of all the negative terms diverges). Then resume putting just enough positive terms to bring the partial sum back up above $beta$. Continue working back and forth like this.
Notice that I said just enough terms at every stage. That ensures that, when you get a partial sum $s$ above $beta$, it won't be too far above $beta$; the difference $s-beta$ will be at most the last term you added, because otherwise you could have stopped adding positive terms sooner. Similarly, when the partial sum goes below $alpha$, the difference will be (in absolute value) at most the (absolute value of the) last term you added.
But your original series converged (conditionally), so the terms approach zero. That means that the amounts by which you overshoot $beta$ and undershoot $alpha$ are eventually arbitrarily small as you perform more and more stages of the process. And that's what those last two lines in Rudin's proof are saying is "clear".
answered Jul 28 '18 at 0:59
Andreas BlassAndreas Blass
50.5k452109
$endgroup$
First, let me make sure it's clear what's happening behind all those formulas. The rearrangement that allegedly works goes like this: First take just enough positive terms from your given series to produce a partial sum $>beta$. (You can do that because the series of all the positive terms diverges.) After that, put just enough negative terms to bring the partial sum down below $alpha$ (possible because the series of all the negative terms diverges). Then resume putting just enough positive terms to bring the partial sum back up above $beta$. Continue working back and forth like this.
Notice that I said just enough terms at every stage. That ensures that, when you get a partial sum $s$ above $beta$, it won't be too far above $beta$; the difference $s-beta$ will be at most the last term you added, because otherwise you could have stopped adding positive terms sooner. Similarly, when the partial sum goes below $alpha$, the difference will be (in absolute value) at most the (absolute value of the) last term you added.
But your original series converged (conditionally), so the terms approach zero. That means that the amounts by which you overshoot $beta$ and undershoot $alpha$ are eventually arbitrarily small as you perform more and more stages of the process. And that's what those last two lines in Rudin's proof are saying is "clear".
answered Jul 28 '18 at 0:59
Andreas BlassAndreas Blass
50.5k452109
answered Jul 28 '18 at 0:59
Andreas BlassAndreas Blass
50.5k452109
answered Jul 28 '18 at 0:59
Andreas BlassAndreas Blass
50.5k452109
answered Jul 28 '18 at 0:59
Andreas BlassAndreas Blass
50.5k452109
50.5k452109
add a comment |
add a comment |
$begingroup$
I will consider the case $beta in mathbb{R}$ here.
Rudin showed that $x_n to beta$ on p.77.
Let $epsilon$ be an arbitrary positive real number.
Then, there exists a natural number $N$ such that $$n geq N implies beta - epsilon < x_n < beta + epsilon.$$
By the construction of ${s'_n}$, the follwoing inequalities hold:
$$beta + epsilon > x_N = s'_{m_1 + k_1 + cdots + m_{N-1} + k_{N-1} + m_N} > s'_{m_1 + k_1 + cdots + m_{N-1} + k_{N-1} + m_N + 1} \> cdots > s'_{m_1 + k_1 + cdots + m_{N-1} + k_{N-1} + m_N + k_N} = y_N < s'_{m_1 + k_1 + cdots + m_{N-1} + k_{N-1} + m_N + k_N+1} < cdots < s'_{m_1 + k_1 + cdots + m_{N-1} + k_{N-1} + m_N + k_N+(m_{N+1}-1)} < s'_{m_1 + k_1 + cdots + m_{N-1} + k_{N-1} + m_N + k_N+m_{N+1}} = x_{N+1} < beta + epsilon.$$
Let $M := m_1 + k_1 + cdots + m_{N-1} + k_{N-1} + m_N$.
The above inequalities say that $s'_n < beta + epsilon$ for all $n in {M, M+1, cdots, M + k_N + m_{N+1}}$.
And we see that the following ineqality holds:
$s'_n < beta + epsilon$ for all $n in {M, M+1, cdots}$.
answered Jan 30 at 9:28
tchappy hatchappy ha
778412
$endgroup$
$begingroup$
Thank you for your answer. I will look at it once I find time :)
$endgroup$
– user370967
Jan 30 at 17:52
add a comment |
$begingroup$
I will consider the case $beta in mathbb{R}$ here.
Rudin showed that $x_n to beta$ on p.77.
Let $epsilon$ be an arbitrary positive real number.
Then, there exists a natural number $N$ such that $$n geq N implies beta - epsilon < x_n < beta + epsilon.$$
By the construction of ${s'_n}$, the follwoing inequalities hold:
$$beta + epsilon > x_N = s'_{m_1 + k_1 + cdots + m_{N-1} + k_{N-1} + m_N} > s'_{m_1 + k_1 + cdots + m_{N-1} + k_{N-1} + m_N + 1} \> cdots > s'_{m_1 + k_1 + cdots + m_{N-1} + k_{N-1} + m_N + k_N} = y_N < s'_{m_1 + k_1 + cdots + m_{N-1} + k_{N-1} + m_N + k_N+1} < cdots < s'_{m_1 + k_1 + cdots + m_{N-1} + k_{N-1} + m_N + k_N+(m_{N+1}-1)} < s'_{m_1 + k_1 + cdots + m_{N-1} + k_{N-1} + m_N + k_N+m_{N+1}} = x_{N+1} < beta + epsilon.$$
Let $M := m_1 + k_1 + cdots + m_{N-1} + k_{N-1} + m_N$.
The above inequalities say that $s'_n < beta + epsilon$ for all $n in {M, M+1, cdots, M + k_N + m_{N+1}}$.
And we see that the following ineqality holds:
$s'_n < beta + epsilon$ for all $n in {M, M+1, cdots}$.
answered Jan 30 at 9:28
tchappy hatchappy ha
778412
$endgroup$
$begingroup$
Thank you for your answer. I will look at it once I find time :)
$endgroup$
– user370967
Jan 30 at 17:52
add a comment |
$begingroup$
I will consider the case $beta in mathbb{R}$ here.
Rudin showed that $x_n to beta$ on p.77.
Let $epsilon$ be an arbitrary positive real number.
Then, there exists a natural number $N$ such that $$n geq N implies beta - epsilon < x_n < beta + epsilon.$$
By the construction of ${s'_n}$, the follwoing inequalities hold:
$$beta + epsilon > x_N = s'_{m_1 + k_1 + cdots + m_{N-1} + k_{N-1} + m_N} > s'_{m_1 + k_1 + cdots + m_{N-1} + k_{N-1} + m_N + 1} \> cdots > s'_{m_1 + k_1 + cdots + m_{N-1} + k_{N-1} + m_N + k_N} = y_N < s'_{m_1 + k_1 + cdots + m_{N-1} + k_{N-1} + m_N + k_N+1} < cdots < s'_{m_1 + k_1 + cdots + m_{N-1} + k_{N-1} + m_N + k_N+(m_{N+1}-1)} < s'_{m_1 + k_1 + cdots + m_{N-1} + k_{N-1} + m_N + k_N+m_{N+1}} = x_{N+1} < beta + epsilon.$$
Let $M := m_1 + k_1 + cdots + m_{N-1} + k_{N-1} + m_N$.
The above inequalities say that $s'_n < beta + epsilon$ for all $n in {M, M+1, cdots, M + k_N + m_{N+1}}$.
And we see that the following ineqality holds:
$s'_n < beta + epsilon$ for all $n in {M, M+1, cdots}$.
answered Jan 30 at 9:28
tchappy hatchappy ha
778412
$endgroup$
I will consider the case $beta in mathbb{R}$ here.
Rudin showed that $x_n to beta$ on p.77.
Let $epsilon$ be an arbitrary positive real number.
Then, there exists a natural number $N$ such that $$n geq N implies beta - epsilon < x_n < beta + epsilon.$$
By the construction of ${s'_n}$, the follwoing inequalities hold:
$$beta + epsilon > x_N = s'_{m_1 + k_1 + cdots + m_{N-1} + k_{N-1} + m_N} > s'_{m_1 + k_1 + cdots + m_{N-1} + k_{N-1} + m_N + 1} \> cdots > s'_{m_1 + k_1 + cdots + m_{N-1} + k_{N-1} + m_N + k_N} = y_N < s'_{m_1 + k_1 + cdots + m_{N-1} + k_{N-1} + m_N + k_N+1} < cdots < s'_{m_1 + k_1 + cdots + m_{N-1} + k_{N-1} + m_N + k_N+(m_{N+1}-1)} < s'_{m_1 + k_1 + cdots + m_{N-1} + k_{N-1} + m_N + k_N+m_{N+1}} = x_{N+1} < beta + epsilon.$$
Let $M := m_1 + k_1 + cdots + m_{N-1} + k_{N-1} + m_N$.
The above inequalities say that $s'_n < beta + epsilon$ for all $n in {M, M+1, cdots, M + k_N + m_{N+1}}$.
And we see that the following ineqality holds:
$s'_n < beta + epsilon$ for all $n in {M, M+1, cdots}$.
answered Jan 30 at 9:28
tchappy hatchappy ha
778412
answered Jan 30 at 9:28
tchappy hatchappy ha
778412
answered Jan 30 at 9:28
tchappy hatchappy ha
778412
answered Jan 30 at 9:28
tchappy hatchappy ha
778412
778412
$begingroup$
Thank you for your answer. I will look at it once I find time :)
$endgroup$
– user370967
Jan 30 at 17:52
add a comment |
$begingroup$
Thank you for your answer. I will look at it once I find time :)
$endgroup$
– user370967
Jan 30 at 17:52
$begingroup$
Thank you for your answer. I will look at it once I find time :)
$endgroup$
– user370967
Jan 30 at 17:52
$begingroup$
Thank you for your answer. I will look at it once I find time :)
$endgroup$
– user370967
Jan 30 at 17:52
add a comment |
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$begingroup$
I'm not sure what you want to number or I'd edit the question myself. You can number an equation as (24) by adding tag{24} at the end of it.
$endgroup$
– saulspatz
Jul 24 '18 at 15:29
$begingroup$
Thanks for your comment. I edited the post.
$endgroup$
– user370967
Jul 24 '18 at 15:34