Rudin exercise 3.8 solutionShowing a series converges absolutely iff it has a uniform boundIf $sum a_n$ is convergent, with $a_n geq 0$ $forall n in mathbbN$, show that if $p > 1$, $sum a_n^p$ converges.Why doesn't this work for Rudin Exercise 3.8Convergence of recursive sequence $a_n+1 =frac 1k left(a_n + fracka_nright)$Two p-norms are not equivalent for different p on $ell_1$Sum of squares converges given sum doesWhat's significant in Abel's Theorem Proof in Baby Rudin?Why is my proof wrong? (Rudin Ch 3 #8) If $sum a_n$ converges, and if $b_n$ is monotonic and bounded, prove that $sum a_n b_n$ converges.Calculus-probabilityProof Verification: A monotonically increasing sequence that is bounded above always has a LUB
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Rudin exercise 3.8 solution
Showing a series converges absolutely iff it has a uniform boundIf $sum a_n$ is convergent, with $a_n geq 0$ $forall n in mathbbN$, show that if $p > 1$, $sum a_n^p$ converges.Why doesn't this work for Rudin Exercise 3.8Convergence of recursive sequence $a_n+1 =frac 1k left(a_n + fracka_nright)$Two p-norms are not equivalent for different p on $ell_1$Sum of squares converges given sum doesWhat's significant in Abel's Theorem Proof in Baby Rudin?Why is my proof wrong? (Rudin Ch 3 #8) If $sum a_n$ converges, and if $b_n$ is monotonic and bounded, prove that $sum a_n b_n$ converges.Calculus-probabilityProof Verification: A monotonically increasing sequence that is bounded above always has a LUB
$begingroup$
Problem . if the sequence $sumlimits_a_n$ converges and $b_n$ is a monotonic bounded sequence then $sumlimits_ a_nb_n$ converges.
Solution. Let's use the identity:
$sumlimits_n=p^q a_nb_n$ = $sumlimits_n=p^q-1 s_n(b_n-b_n-1)+s_qb_q-s_p-1b_p$, where $s_n=sumlimits_k=1^n a_n$. Then $|sumlimits_n=p^q-1 s_n(b_n-b_n-1)|<=|sumlimits_n=p^q-1 s_n|B$, where B is the difference of upper and lower bound of the sequence $b_n$. since $a_n$ is convergent we can make $|sumlimits_n=p^q-1 s_n|$ as small as we want and therefore we can make this $|sumlimits_n=p^q-1 s_n(b_n-b_n-1)|$ as small as we want like $x/3$. Then let's take the absolute value of another term of the right side of the identity.
$|s_qb_q|<=|s_q||b_up|$, where $b_up$ is the upper bound of the sequence and the same argument here too. we can make this term as small as we want. The same goes to the absolute value of the third term on the right side of the identity. Finally $|sumlimits_n=p^q a_nb_n|<=|sumlimits_n=p^q-1 s_n(b_n-b_n-1)|+|s_qb_q|+|s_p-1b_p|$. We can find $p$ and $q$ to make the right side small enough, completing the proof. Is this right? I am not satisfied what I have written here, I should have stated better in my opinion but the general idea of the proof is shown I suppose.
real-analysis proof-verification
edited Mar 15 at 18:05
Alex Provost
15.6k22351
asked Mar 15 at 17:59
shota kobakhidzeshota kobakhidze
115
$endgroup$
add a comment |
$begingroup$
Problem . if the sequence $sumlimits_a_n$ converges and $b_n$ is a monotonic bounded sequence then $sumlimits_ a_nb_n$ converges.
Solution. Let's use the identity:
$sumlimits_n=p^q a_nb_n$ = $sumlimits_n=p^q-1 s_n(b_n-b_n-1)+s_qb_q-s_p-1b_p$, where $s_n=sumlimits_k=1^n a_n$. Then $|sumlimits_n=p^q-1 s_n(b_n-b_n-1)|<=|sumlimits_n=p^q-1 s_n|B$, where B is the difference of upper and lower bound of the sequence $b_n$. since $a_n$ is convergent we can make $|sumlimits_n=p^q-1 s_n|$ as small as we want and therefore we can make this $|sumlimits_n=p^q-1 s_n(b_n-b_n-1)|$ as small as we want like $x/3$. Then let's take the absolute value of another term of the right side of the identity.
$|s_qb_q|<=|s_q||b_up|$, where $b_up$ is the upper bound of the sequence and the same argument here too. we can make this term as small as we want. The same goes to the absolute value of the third term on the right side of the identity. Finally $|sumlimits_n=p^q a_nb_n|<=|sumlimits_n=p^q-1 s_n(b_n-b_n-1)|+|s_qb_q|+|s_p-1b_p|$. We can find $p$ and $q$ to make the right side small enough, completing the proof. Is this right? I am not satisfied what I have written here, I should have stated better in my opinion but the general idea of the proof is shown I suppose.
real-analysis proof-verification
edited Mar 15 at 18:05
Alex Provost
15.6k22351
asked Mar 15 at 17:59
shota kobakhidzeshota kobakhidze
115
$endgroup$
$begingroup$
yeah, "making as small as we want" is a bad idea, but I can say that I can bound that sum by taking appropriate $p$ and $q$.
$endgroup$
– shota kobakhidze
Mar 15 at 18:28
add a comment |
$begingroup$
Problem . if the sequence $sumlimits_a_n$ converges and $b_n$ is a monotonic bounded sequence then $sumlimits_ a_nb_n$ converges.
Solution. Let's use the identity:
$sumlimits_n=p^q a_nb_n$ = $sumlimits_n=p^q-1 s_n(b_n-b_n-1)+s_qb_q-s_p-1b_p$, where $s_n=sumlimits_k=1^n a_n$. Then $|sumlimits_n=p^q-1 s_n(b_n-b_n-1)|<=|sumlimits_n=p^q-1 s_n|B$, where B is the difference of upper and lower bound of the sequence $b_n$. since $a_n$ is convergent we can make $|sumlimits_n=p^q-1 s_n|$ as small as we want and therefore we can make this $|sumlimits_n=p^q-1 s_n(b_n-b_n-1)|$ as small as we want like $x/3$. Then let's take the absolute value of another term of the right side of the identity.
$|s_qb_q|<=|s_q||b_up|$, where $b_up$ is the upper bound of the sequence and the same argument here too. we can make this term as small as we want. The same goes to the absolute value of the third term on the right side of the identity. Finally $|sumlimits_n=p^q a_nb_n|<=|sumlimits_n=p^q-1 s_n(b_n-b_n-1)|+|s_qb_q|+|s_p-1b_p|$. We can find $p$ and $q$ to make the right side small enough, completing the proof. Is this right? I am not satisfied what I have written here, I should have stated better in my opinion but the general idea of the proof is shown I suppose.
real-analysis proof-verification
edited Mar 15 at 18:05
Alex Provost
15.6k22351
asked Mar 15 at 17:59
shota kobakhidzeshota kobakhidze
115
$endgroup$
Problem . if the sequence $sumlimits_a_n$ converges and $b_n$ is a monotonic bounded sequence then $sumlimits_ a_nb_n$ converges.
Solution. Let's use the identity:
$sumlimits_n=p^q a_nb_n$ = $sumlimits_n=p^q-1 s_n(b_n-b_n-1)+s_qb_q-s_p-1b_p$, where $s_n=sumlimits_k=1^n a_n$. Then $|sumlimits_n=p^q-1 s_n(b_n-b_n-1)|<=|sumlimits_n=p^q-1 s_n|B$, where B is the difference of upper and lower bound of the sequence $b_n$. since $a_n$ is convergent we can make $|sumlimits_n=p^q-1 s_n|$ as small as we want and therefore we can make this $|sumlimits_n=p^q-1 s_n(b_n-b_n-1)|$ as small as we want like $x/3$. Then let's take the absolute value of another term of the right side of the identity.
$|s_qb_q|<=|s_q||b_up|$, where $b_up$ is the upper bound of the sequence and the same argument here too. we can make this term as small as we want. The same goes to the absolute value of the third term on the right side of the identity. Finally $|sumlimits_n=p^q a_nb_n|<=|sumlimits_n=p^q-1 s_n(b_n-b_n-1)|+|s_qb_q|+|s_p-1b_p|$. We can find $p$ and $q$ to make the right side small enough, completing the proof. Is this right? I am not satisfied what I have written here, I should have stated better in my opinion but the general idea of the proof is shown I suppose.
real-analysis proof-verification
real-analysis proof-verification
edited Mar 15 at 18:05
Alex Provost
15.6k22351
asked Mar 15 at 17:59
shota kobakhidzeshota kobakhidze
115
edited Mar 15 at 18:05
Alex Provost
15.6k22351
asked Mar 15 at 17:59
shota kobakhidzeshota kobakhidze
115
edited Mar 15 at 18:05
Alex Provost
15.6k22351
edited Mar 15 at 18:05
Alex Provost
15.6k22351
edited Mar 15 at 18:05
Alex Provost
15.6k22351
15.6k22351
asked Mar 15 at 17:59
shota kobakhidzeshota kobakhidze
115
asked Mar 15 at 17:59
shota kobakhidzeshota kobakhidze
115
asked Mar 15 at 17:59
shota kobakhidzeshota kobakhidze
115
115
$begingroup$
yeah, "making as small as we want" is a bad idea, but I can say that I can bound that sum by taking appropriate $p$ and $q$.
$endgroup$
– shota kobakhidze
Mar 15 at 18:28
add a comment |
$begingroup$
yeah, "making as small as we want" is a bad idea, but I can say that I can bound that sum by taking appropriate $p$ and $q$.
$endgroup$
– shota kobakhidze
Mar 15 at 18:28
$begingroup$
yeah, "making as small as we want" is a bad idea, but I can say that I can bound that sum by taking appropriate $p$ and $q$.
$endgroup$
– shota kobakhidze
Mar 15 at 18:28
$begingroup$
yeah, "making as small as we want" is a bad idea, but I can say that I can bound that sum by taking appropriate $p$ and $q$.
$endgroup$
– shota kobakhidze
Mar 15 at 18:28
add a comment |
2 Answers
2
active
oldest
votes
$begingroup$
Assume that $b_n$ is increasing. If $s_n=sum_k=1^n a_k$, then
$$
sum_k=1^n a_kb_k=sum_k=1^n (s_k-s_k-1)b_k=sum_k=1^n s_kb_k-sum_k=2^n s_k-1b_k=s_nb_n+
sum_k=2^ns_k-1(b_k-1-b_k).
$$
Now $s_nb_n$ is convergent, as a product of convergent sequences, while the series
$sum s_n-1(b_n-1-b_n)$
is dominated by the series $sum M(b_n-b_n-1)$, where $M=sup_ninmathbb N|s_n|$, and since $b_n-b_n-1$ is absolutely summable,
so is $sum s_n-1(b_n-1-b_n)$.
answered Mar 15 at 19:03
Yiorgos S. SmyrlisYiorgos S. Smyrlis
63.6k1385165
$endgroup$
$begingroup$
yeah, it looks like a similar idea, stated brilliantly.
$endgroup$
– shota kobakhidze
Mar 15 at 19:11
add a comment |
$begingroup$
The general idea is correct. The expression “the upper bound of the sequence” is not appropriate (there are infinitely many upper bounds). I think that it would be better to say that $b_up$ is an upper bound of the sequence $bigl(lvert b_nrvertbigr)_ninmathbb N$. With this choice, $b_up$ is non-negative. Also, with this choice you have$$leftlvertsum_n=p^q-1s_n(b_n-b_n-1)rightrvertleqslant2leftlvertsum_n=p^q-1s_nrightrvert b_up.$$Again, the expression “the difference of upper and lower bound of the sequence $b_n$” is not appropriate.
edited Mar 15 at 19:16
J. W. Tanner
3,6631320
answered Mar 15 at 18:29
José Carlos SantosJosé Carlos Santos
169k23132238
$endgroup$
$begingroup$
yeah, understood. Thanks for the response. By the way, saying that "we can make it as small as we want" is incorrect right? It would be more correct to say "bounded". Then the whole sum becomes "bounded" and not "as small as I wanted".
$endgroup$
– shota kobakhidze
Mar 15 at 18:38
$begingroup$
Oops! You are right. What you shoud use at that point is that $(s_n)_ninmathbb N$ is a boounded sequence and that the series $sum_n=1^inftylvert b_n-b_n+1rvert$ converges (this follows from the fact that $(b_n)_ninmathbb N$ is a monotonic and bounded sequence).
$endgroup$
– José Carlos Santos
Mar 15 at 18:47
$begingroup$
By the way, is it necessary for $b_n$ to be monotonic? cause I haven't used that fact, I just used the boundedness.
$endgroup$
– shota kobakhidze
Mar 15 at 23:16
$begingroup$
See my previous comment. Without that sequence being monotonic, how do you know that the series $sum_n=1^inftylvert b_n-b_n+1rvert$ converges?
$endgroup$
– José Carlos Santos
Mar 15 at 23:33
$begingroup$
yes, I see. It was my bad. Thanks :)
$endgroup$
– shota kobakhidze
Mar 16 at 0:00
add a comment |
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2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
Assume that $b_n$ is increasing. If $s_n=sum_k=1^n a_k$, then
$$
sum_k=1^n a_kb_k=sum_k=1^n (s_k-s_k-1)b_k=sum_k=1^n s_kb_k-sum_k=2^n s_k-1b_k=s_nb_n+
sum_k=2^ns_k-1(b_k-1-b_k).
$$
Now $s_nb_n$ is convergent, as a product of convergent sequences, while the series
$sum s_n-1(b_n-1-b_n)$
is dominated by the series $sum M(b_n-b_n-1)$, where $M=sup_ninmathbb N|s_n|$, and since $b_n-b_n-1$ is absolutely summable,
so is $sum s_n-1(b_n-1-b_n)$.
answered Mar 15 at 19:03
Yiorgos S. SmyrlisYiorgos S. Smyrlis
63.6k1385165
$endgroup$
$begingroup$
yeah, it looks like a similar idea, stated brilliantly.
$endgroup$
– shota kobakhidze
Mar 15 at 19:11
add a comment |
$begingroup$
Assume that $b_n$ is increasing. If $s_n=sum_k=1^n a_k$, then
$$
sum_k=1^n a_kb_k=sum_k=1^n (s_k-s_k-1)b_k=sum_k=1^n s_kb_k-sum_k=2^n s_k-1b_k=s_nb_n+
sum_k=2^ns_k-1(b_k-1-b_k).
$$
Now $s_nb_n$ is convergent, as a product of convergent sequences, while the series
$sum s_n-1(b_n-1-b_n)$
is dominated by the series $sum M(b_n-b_n-1)$, where $M=sup_ninmathbb N|s_n|$, and since $b_n-b_n-1$ is absolutely summable,
so is $sum s_n-1(b_n-1-b_n)$.
answered Mar 15 at 19:03
Yiorgos S. SmyrlisYiorgos S. Smyrlis
63.6k1385165
$endgroup$
$begingroup$
yeah, it looks like a similar idea, stated brilliantly.
$endgroup$
– shota kobakhidze
Mar 15 at 19:11
add a comment |
$begingroup$
Assume that $b_n$ is increasing. If $s_n=sum_k=1^n a_k$, then
$$
sum_k=1^n a_kb_k=sum_k=1^n (s_k-s_k-1)b_k=sum_k=1^n s_kb_k-sum_k=2^n s_k-1b_k=s_nb_n+
sum_k=2^ns_k-1(b_k-1-b_k).
$$
Now $s_nb_n$ is convergent, as a product of convergent sequences, while the series
$sum s_n-1(b_n-1-b_n)$
is dominated by the series $sum M(b_n-b_n-1)$, where $M=sup_ninmathbb N|s_n|$, and since $b_n-b_n-1$ is absolutely summable,
so is $sum s_n-1(b_n-1-b_n)$.
answered Mar 15 at 19:03
Yiorgos S. SmyrlisYiorgos S. Smyrlis
63.6k1385165
$endgroup$
Assume that $b_n$ is increasing. If $s_n=sum_k=1^n a_k$, then
$$
sum_k=1^n a_kb_k=sum_k=1^n (s_k-s_k-1)b_k=sum_k=1^n s_kb_k-sum_k=2^n s_k-1b_k=s_nb_n+
sum_k=2^ns_k-1(b_k-1-b_k).
$$
Now $s_nb_n$ is convergent, as a product of convergent sequences, while the series
$sum s_n-1(b_n-1-b_n)$
is dominated by the series $sum M(b_n-b_n-1)$, where $M=sup_ninmathbb N|s_n|$, and since $b_n-b_n-1$ is absolutely summable,
so is $sum s_n-1(b_n-1-b_n)$.
answered Mar 15 at 19:03
Yiorgos S. SmyrlisYiorgos S. Smyrlis
63.6k1385165
answered Mar 15 at 19:03
Yiorgos S. SmyrlisYiorgos S. Smyrlis
63.6k1385165
answered Mar 15 at 19:03
Yiorgos S. SmyrlisYiorgos S. Smyrlis
63.6k1385165
answered Mar 15 at 19:03
Yiorgos S. SmyrlisYiorgos S. Smyrlis
63.6k1385165
63.6k1385165
$begingroup$
yeah, it looks like a similar idea, stated brilliantly.
$endgroup$
– shota kobakhidze
Mar 15 at 19:11
add a comment |
$begingroup$
yeah, it looks like a similar idea, stated brilliantly.
$endgroup$
– shota kobakhidze
Mar 15 at 19:11
$begingroup$
yeah, it looks like a similar idea, stated brilliantly.
$endgroup$
– shota kobakhidze
Mar 15 at 19:11
$begingroup$
yeah, it looks like a similar idea, stated brilliantly.
$endgroup$
– shota kobakhidze
Mar 15 at 19:11
add a comment |
$begingroup$
The general idea is correct. The expression “the upper bound of the sequence” is not appropriate (there are infinitely many upper bounds). I think that it would be better to say that $b_up$ is an upper bound of the sequence $bigl(lvert b_nrvertbigr)_ninmathbb N$. With this choice, $b_up$ is non-negative. Also, with this choice you have$$leftlvertsum_n=p^q-1s_n(b_n-b_n-1)rightrvertleqslant2leftlvertsum_n=p^q-1s_nrightrvert b_up.$$Again, the expression “the difference of upper and lower bound of the sequence $b_n$” is not appropriate.
edited Mar 15 at 19:16
J. W. Tanner
3,6631320
answered Mar 15 at 18:29
José Carlos SantosJosé Carlos Santos
169k23132238
$endgroup$
$begingroup$
yeah, understood. Thanks for the response. By the way, saying that "we can make it as small as we want" is incorrect right? It would be more correct to say "bounded". Then the whole sum becomes "bounded" and not "as small as I wanted".
$endgroup$
– shota kobakhidze
Mar 15 at 18:38
$begingroup$
Oops! You are right. What you shoud use at that point is that $(s_n)_ninmathbb N$ is a boounded sequence and that the series $sum_n=1^inftylvert b_n-b_n+1rvert$ converges (this follows from the fact that $(b_n)_ninmathbb N$ is a monotonic and bounded sequence).
$endgroup$
– José Carlos Santos
Mar 15 at 18:47
$begingroup$
By the way, is it necessary for $b_n$ to be monotonic? cause I haven't used that fact, I just used the boundedness.
$endgroup$
– shota kobakhidze
Mar 15 at 23:16
$begingroup$
See my previous comment. Without that sequence being monotonic, how do you know that the series $sum_n=1^inftylvert b_n-b_n+1rvert$ converges?
$endgroup$
– José Carlos Santos
Mar 15 at 23:33
$begingroup$
yes, I see. It was my bad. Thanks :)
$endgroup$
– shota kobakhidze
Mar 16 at 0:00
add a comment |
$begingroup$
The general idea is correct. The expression “the upper bound of the sequence” is not appropriate (there are infinitely many upper bounds). I think that it would be better to say that $b_up$ is an upper bound of the sequence $bigl(lvert b_nrvertbigr)_ninmathbb N$. With this choice, $b_up$ is non-negative. Also, with this choice you have$$leftlvertsum_n=p^q-1s_n(b_n-b_n-1)rightrvertleqslant2leftlvertsum_n=p^q-1s_nrightrvert b_up.$$Again, the expression “the difference of upper and lower bound of the sequence $b_n$” is not appropriate.
edited Mar 15 at 19:16
J. W. Tanner
3,6631320
answered Mar 15 at 18:29
José Carlos SantosJosé Carlos Santos
169k23132238
$endgroup$
$begingroup$
yeah, understood. Thanks for the response. By the way, saying that "we can make it as small as we want" is incorrect right? It would be more correct to say "bounded". Then the whole sum becomes "bounded" and not "as small as I wanted".
$endgroup$
– shota kobakhidze
Mar 15 at 18:38
$begingroup$
Oops! You are right. What you shoud use at that point is that $(s_n)_ninmathbb N$ is a boounded sequence and that the series $sum_n=1^inftylvert b_n-b_n+1rvert$ converges (this follows from the fact that $(b_n)_ninmathbb N$ is a monotonic and bounded sequence).
$endgroup$
– José Carlos Santos
Mar 15 at 18:47
$begingroup$
By the way, is it necessary for $b_n$ to be monotonic? cause I haven't used that fact, I just used the boundedness.
$endgroup$
– shota kobakhidze
Mar 15 at 23:16
$begingroup$
See my previous comment. Without that sequence being monotonic, how do you know that the series $sum_n=1^inftylvert b_n-b_n+1rvert$ converges?
$endgroup$
– José Carlos Santos
Mar 15 at 23:33
$begingroup$
yes, I see. It was my bad. Thanks :)
$endgroup$
– shota kobakhidze
Mar 16 at 0:00
add a comment |
$begingroup$
The general idea is correct. The expression “the upper bound of the sequence” is not appropriate (there are infinitely many upper bounds). I think that it would be better to say that $b_up$ is an upper bound of the sequence $bigl(lvert b_nrvertbigr)_ninmathbb N$. With this choice, $b_up$ is non-negative. Also, with this choice you have$$leftlvertsum_n=p^q-1s_n(b_n-b_n-1)rightrvertleqslant2leftlvertsum_n=p^q-1s_nrightrvert b_up.$$Again, the expression “the difference of upper and lower bound of the sequence $b_n$” is not appropriate.
edited Mar 15 at 19:16
J. W. Tanner
3,6631320
answered Mar 15 at 18:29
José Carlos SantosJosé Carlos Santos
169k23132238
$endgroup$
The general idea is correct. The expression “the upper bound of the sequence” is not appropriate (there are infinitely many upper bounds). I think that it would be better to say that $b_up$ is an upper bound of the sequence $bigl(lvert b_nrvertbigr)_ninmathbb N$. With this choice, $b_up$ is non-negative. Also, with this choice you have$$leftlvertsum_n=p^q-1s_n(b_n-b_n-1)rightrvertleqslant2leftlvertsum_n=p^q-1s_nrightrvert b_up.$$Again, the expression “the difference of upper and lower bound of the sequence $b_n$” is not appropriate.
edited Mar 15 at 19:16
J. W. Tanner
3,6631320
answered Mar 15 at 18:29
José Carlos SantosJosé Carlos Santos
169k23132238
edited Mar 15 at 19:16
J. W. Tanner
3,6631320
edited Mar 15 at 19:16
J. W. Tanner
3,6631320
edited Mar 15 at 19:16
J. W. Tanner
3,6631320
3,6631320
answered Mar 15 at 18:29
José Carlos SantosJosé Carlos Santos
169k23132238
answered Mar 15 at 18:29
José Carlos SantosJosé Carlos Santos
169k23132238
answered Mar 15 at 18:29
José Carlos SantosJosé Carlos Santos
169k23132238
169k23132238
$begingroup$
yeah, understood. Thanks for the response. By the way, saying that "we can make it as small as we want" is incorrect right? It would be more correct to say "bounded". Then the whole sum becomes "bounded" and not "as small as I wanted".
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– shota kobakhidze
Mar 15 at 18:38
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Oops! You are right. What you shoud use at that point is that $(s_n)_ninmathbb N$ is a boounded sequence and that the series $sum_n=1^inftylvert b_n-b_n+1rvert$ converges (this follows from the fact that $(b_n)_ninmathbb N$ is a monotonic and bounded sequence).
$endgroup$
– José Carlos Santos
Mar 15 at 18:47
$begingroup$
By the way, is it necessary for $b_n$ to be monotonic? cause I haven't used that fact, I just used the boundedness.
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– shota kobakhidze
Mar 15 at 23:16
$begingroup$
See my previous comment. Without that sequence being monotonic, how do you know that the series $sum_n=1^inftylvert b_n-b_n+1rvert$ converges?
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– José Carlos Santos
Mar 15 at 23:33
$begingroup$
yes, I see. It was my bad. Thanks :)
$endgroup$
– shota kobakhidze
Mar 16 at 0:00
add a comment |
$begingroup$
yeah, understood. Thanks for the response. By the way, saying that "we can make it as small as we want" is incorrect right? It would be more correct to say "bounded". Then the whole sum becomes "bounded" and not "as small as I wanted".
$endgroup$
– shota kobakhidze
Mar 15 at 18:38
$begingroup$
Oops! You are right. What you shoud use at that point is that $(s_n)_ninmathbb N$ is a boounded sequence and that the series $sum_n=1^inftylvert b_n-b_n+1rvert$ converges (this follows from the fact that $(b_n)_ninmathbb N$ is a monotonic and bounded sequence).
$endgroup$
– José Carlos Santos
Mar 15 at 18:47
$begingroup$
By the way, is it necessary for $b_n$ to be monotonic? cause I haven't used that fact, I just used the boundedness.
$endgroup$
– shota kobakhidze
Mar 15 at 23:16
$begingroup$
See my previous comment. Without that sequence being monotonic, how do you know that the series $sum_n=1^inftylvert b_n-b_n+1rvert$ converges?
$endgroup$
– José Carlos Santos
Mar 15 at 23:33
$begingroup$
yes, I see. It was my bad. Thanks :)
$endgroup$
– shota kobakhidze
Mar 16 at 0:00
$begingroup$
yeah, understood. Thanks for the response. By the way, saying that "we can make it as small as we want" is incorrect right? It would be more correct to say "bounded". Then the whole sum becomes "bounded" and not "as small as I wanted".
$endgroup$
– shota kobakhidze
Mar 15 at 18:38
$begingroup$
yeah, understood. Thanks for the response. By the way, saying that "we can make it as small as we want" is incorrect right? It would be more correct to say "bounded". Then the whole sum becomes "bounded" and not "as small as I wanted".
$endgroup$
– shota kobakhidze
Mar 15 at 18:38
$begingroup$
Oops! You are right. What you shoud use at that point is that $(s_n)_ninmathbb N$ is a boounded sequence and that the series $sum_n=1^inftylvert b_n-b_n+1rvert$ converges (this follows from the fact that $(b_n)_ninmathbb N$ is a monotonic and bounded sequence).
$endgroup$
– José Carlos Santos
Mar 15 at 18:47
$begingroup$
Oops! You are right. What you shoud use at that point is that $(s_n)_ninmathbb N$ is a boounded sequence and that the series $sum_n=1^inftylvert b_n-b_n+1rvert$ converges (this follows from the fact that $(b_n)_ninmathbb N$ is a monotonic and bounded sequence).
$endgroup$
– José Carlos Santos
Mar 15 at 18:47
$begingroup$
By the way, is it necessary for $b_n$ to be monotonic? cause I haven't used that fact, I just used the boundedness.
$endgroup$
– shota kobakhidze
Mar 15 at 23:16
$begingroup$
By the way, is it necessary for $b_n$ to be monotonic? cause I haven't used that fact, I just used the boundedness.
$endgroup$
– shota kobakhidze
Mar 15 at 23:16
$begingroup$
See my previous comment. Without that sequence being monotonic, how do you know that the series $sum_n=1^inftylvert b_n-b_n+1rvert$ converges?
$endgroup$
– José Carlos Santos
Mar 15 at 23:33
$begingroup$
See my previous comment. Without that sequence being monotonic, how do you know that the series $sum_n=1^inftylvert b_n-b_n+1rvert$ converges?
$endgroup$
– José Carlos Santos
Mar 15 at 23:33
$begingroup$
yes, I see. It was my bad. Thanks :)
$endgroup$
– shota kobakhidze
Mar 16 at 0:00
$begingroup$
yes, I see. It was my bad. Thanks :)
$endgroup$
– shota kobakhidze
Mar 16 at 0:00
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$begingroup$
yeah, "making as small as we want" is a bad idea, but I can say that I can bound that sum by taking appropriate $p$ and $q$.
$endgroup$
– shota kobakhidze
Mar 15 at 18:28