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A series that converges but the square of the terms of the series diverges
Clash Royale CLAN TAG#URR8PPP
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I have to give an example of a convergent series $sum a_n$ for which $sum a_n^2 $ diverges.
I think that such a series cannot exist because if $sum a_n$ converges absolutely then $sum a_n^2 $ will always converge right?
real-analysis sequences-and-series
asked 21 mins ago
Snop D.
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I have to give an example of a convergent series $sum a_n$ for which $sum a_n^2 $ diverges.
I think that such a series cannot exist because if $sum a_n$ converges absolutely then $sum a_n^2 $ will always converge right?
real-analysis sequences-and-series
asked 21 mins ago
Snop D.
61
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Snop D. is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
The implication you said after "because" is true but it's still possible that $sum_n a_n$ converges but not absolutely and $sum_n a_n^2$ does not converge (absolutely nor by any means).
– Saucy O'Path
19 mins ago
@max_zorn That's if $sum a_n$ converges absolutely.
– Robert Israel
15 mins ago
Yes, you are right Robert, sorry!
– max_zorn
12 mins ago
add a comment |Â
up vote
1
down vote
favorite
up vote
1
down vote
favorite
I have to give an example of a convergent series $sum a_n$ for which $sum a_n^2 $ diverges.
I think that such a series cannot exist because if $sum a_n$ converges absolutely then $sum a_n^2 $ will always converge right?
real-analysis sequences-and-series
asked 21 mins ago
Snop D.
61
New contributor
Snop D. is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
I have to give an example of a convergent series $sum a_n$ for which $sum a_n^2 $ diverges.
I think that such a series cannot exist because if $sum a_n$ converges absolutely then $sum a_n^2 $ will always converge right?
real-analysis sequences-and-series
real-analysis sequences-and-series
asked 21 mins ago
Snop D.
61
New contributor
Snop D. is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
asked 21 mins ago
Snop D.
61
New contributor
Snop D. is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
asked 21 mins ago
Snop D.
61
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Snop D. is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
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asked 21 mins ago
Snop D.
61
asked 21 mins ago
Snop D.
61
61
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Snop D. is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
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New contributor
Snop D. is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
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The implication you said after "because" is true but it's still possible that $sum_n a_n$ converges but not absolutely and $sum_n a_n^2$ does not converge (absolutely nor by any means).
– Saucy O'Path
19 mins ago
@max_zorn That's if $sum a_n$ converges absolutely.
– Robert Israel
15 mins ago
Yes, you are right Robert, sorry!
– max_zorn
12 mins ago
add a comment |Â
The implication you said after "because" is true but it's still possible that $sum_n a_n$ converges but not absolutely and $sum_n a_n^2$ does not converge (absolutely nor by any means).
– Saucy O'Path
19 mins ago
@max_zorn That's if $sum a_n$ converges absolutely.
– Robert Israel
15 mins ago
Yes, you are right Robert, sorry!
– max_zorn
12 mins ago
The implication you said after "because" is true but it's still possible that $sum_n a_n$ converges but not absolutely and $sum_n a_n^2$ does not converge (absolutely nor by any means).
– Saucy O'Path
19 mins ago
The implication you said after "because" is true but it's still possible that $sum_n a_n$ converges but not absolutely and $sum_n a_n^2$ does not converge (absolutely nor by any means).
– Saucy O'Path
19 mins ago
@max_zorn That's if $sum a_n$ converges absolutely.
– Robert Israel
15 mins ago
@max_zorn That's if $sum a_n$ converges absolutely.
– Robert Israel
15 mins ago
Yes, you are right Robert, sorry!
– max_zorn
12 mins ago
Yes, you are right Robert, sorry!
– max_zorn
12 mins ago
add a comment |Â
3 Answers
3
active
oldest
votes
up vote
3
down vote
Try $$sum_n=1^infty frac(-1)^nsqrtn$$
answered 18 mins ago
Robert Israel
312k23204450
add a comment |Â
up vote
2
down vote
The alternating series test gives a wealth of examples. Take
$$
a_n=(-1)^n/sqrtn
$$
for example.
answered 18 mins ago
Foobaz John
19.1k41247
You beat me to it by $14$ seconds.
– Robert Israel
17 mins ago
add a comment |Â
up vote
0
down vote
The simplest example I have in mind is :
$$sum_n=1^+inftyfrac(-1)^nsqrtn $$
Beware that convergent (CV) and absolutely convergent (ACV) can be very different.
Indeed, if $sum_n=1^+infty a_n $ is ACV then, $sum_n=1^+infty a_n^2$ is also ACV. To prove it, you can simply notice that, since $lim a_n=0$ then for $n$ big enough, $|a_n^2|<|a_n|$.
You can also ask yourself a more general question :
What are the function $f:mathbb R rightarrow mathbb R$ such that for all
$sum a_n$ CV (resp. ACV), $sum f(a_n)$ is CV (resp. ACV).
It is a (difficult) exercise to show that for $f:mathbb Rrightarrow mathbb R$,
$$sum a_n ~CV Rightarrow sum f(a_n) ~CV quad textiff quad exists eta>0,exists lambdain mathbb R,forall xin ]-eta,eta[, quad f(x)=lambda x $$
$$sum a_n ~CV Rightarrow sum f(a_n) ~ACV quad textiff quad exists eta>0,forall xin ]-eta,eta[, quad f(x)=0 quadquadquad~~$$
$$sum a_n ~ACV Rightarrow sum f(a_n) ~ACV quad textiff quadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquad$$ $$quad f(0)=0 text and exists eta>0,exists M >0,forall xin ]-eta,eta[, |f(x)|leq M |x| $$
$$sum a_n ACV Rightarrow sum f(a_n) ~CV quad textiff quad sum a_n ~ACV Rightarrow sum f(a_n) ~ACV~~~~~~~~~~~~~~ $$
answered 19 mins ago
Can I play with Mathness
1443
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Can I play with Mathness is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
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1
$sum_n a_n^2$ converges for this one.
– Robert Israel
17 mins ago
oups I answer too fast ^^
– Can I play with Mathness
14 mins ago
Can you correct it? Want to upvote the distinction between ACV and CV but can't bring myself to do it with the error
– Oscar Lanzi
5 mins ago
corrected, I also added a more general statement.
– Can I play with Mathness
3 mins ago
add a comment |Â
3 Answers
3
active
oldest
votes
3 Answers
3
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
3
down vote
Try $$sum_n=1^infty frac(-1)^nsqrtn$$
answered 18 mins ago
Robert Israel
312k23204450
add a comment |Â
up vote
3
down vote
Try $$sum_n=1^infty frac(-1)^nsqrtn$$
answered 18 mins ago
Robert Israel
312k23204450
add a comment |Â
up vote
3
down vote
up vote
3
down vote
Try $$sum_n=1^infty frac(-1)^nsqrtn$$
answered 18 mins ago
Robert Israel
312k23204450
Try $$sum_n=1^infty frac(-1)^nsqrtn$$
answered 18 mins ago
Robert Israel
312k23204450
answered 18 mins ago
Robert Israel
312k23204450
answered 18 mins ago
Robert Israel
312k23204450
answered 18 mins ago
Robert Israel
312k23204450
312k23204450
add a comment |Â
add a comment |Â
up vote
2
down vote
The alternating series test gives a wealth of examples. Take
$$
a_n=(-1)^n/sqrtn
$$
for example.
answered 18 mins ago
Foobaz John
19.1k41247
You beat me to it by $14$ seconds.
– Robert Israel
17 mins ago
add a comment |Â
up vote
2
down vote
The alternating series test gives a wealth of examples. Take
$$
a_n=(-1)^n/sqrtn
$$
for example.
answered 18 mins ago
Foobaz John
19.1k41247
You beat me to it by $14$ seconds.
– Robert Israel
17 mins ago
add a comment |Â
up vote
2
down vote
up vote
2
down vote
The alternating series test gives a wealth of examples. Take
$$
a_n=(-1)^n/sqrtn
$$
for example.
answered 18 mins ago
Foobaz John
19.1k41247
The alternating series test gives a wealth of examples. Take
$$
a_n=(-1)^n/sqrtn
$$
for example.
answered 18 mins ago
Foobaz John
19.1k41247
answered 18 mins ago
Foobaz John
19.1k41247
answered 18 mins ago
Foobaz John
19.1k41247
answered 18 mins ago
Foobaz John
19.1k41247
19.1k41247
You beat me to it by $14$ seconds.
– Robert Israel
17 mins ago
add a comment |Â
You beat me to it by $14$ seconds.
– Robert Israel
17 mins ago
You beat me to it by $14$ seconds.
– Robert Israel
17 mins ago
You beat me to it by $14$ seconds.
– Robert Israel
17 mins ago
add a comment |Â
up vote
0
down vote
The simplest example I have in mind is :
$$sum_n=1^+inftyfrac(-1)^nsqrtn $$
Beware that convergent (CV) and absolutely convergent (ACV) can be very different.
Indeed, if $sum_n=1^+infty a_n $ is ACV then, $sum_n=1^+infty a_n^2$ is also ACV. To prove it, you can simply notice that, since $lim a_n=0$ then for $n$ big enough, $|a_n^2|<|a_n|$.
You can also ask yourself a more general question :
What are the function $f:mathbb R rightarrow mathbb R$ such that for all
$sum a_n$ CV (resp. ACV), $sum f(a_n)$ is CV (resp. ACV).
It is a (difficult) exercise to show that for $f:mathbb Rrightarrow mathbb R$,
$$sum a_n ~CV Rightarrow sum f(a_n) ~CV quad textiff quad exists eta>0,exists lambdain mathbb R,forall xin ]-eta,eta[, quad f(x)=lambda x $$
$$sum a_n ~CV Rightarrow sum f(a_n) ~ACV quad textiff quad exists eta>0,forall xin ]-eta,eta[, quad f(x)=0 quadquadquad~~$$
$$sum a_n ~ACV Rightarrow sum f(a_n) ~ACV quad textiff quadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquad$$ $$quad f(0)=0 text and exists eta>0,exists M >0,forall xin ]-eta,eta[, |f(x)|leq M |x| $$
$$sum a_n ACV Rightarrow sum f(a_n) ~CV quad textiff quad sum a_n ~ACV Rightarrow sum f(a_n) ~ACV~~~~~~~~~~~~~~ $$
answered 19 mins ago
Can I play with Mathness
1443
New contributor
Can I play with Mathness is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
1
$sum_n a_n^2$ converges for this one.
– Robert Israel
17 mins ago
oups I answer too fast ^^
– Can I play with Mathness
14 mins ago
Can you correct it? Want to upvote the distinction between ACV and CV but can't bring myself to do it with the error
– Oscar Lanzi
5 mins ago
corrected, I also added a more general statement.
– Can I play with Mathness
3 mins ago
add a comment |Â
up vote
0
down vote
The simplest example I have in mind is :
$$sum_n=1^+inftyfrac(-1)^nsqrtn $$
Beware that convergent (CV) and absolutely convergent (ACV) can be very different.
Indeed, if $sum_n=1^+infty a_n $ is ACV then, $sum_n=1^+infty a_n^2$ is also ACV. To prove it, you can simply notice that, since $lim a_n=0$ then for $n$ big enough, $|a_n^2|<|a_n|$.
You can also ask yourself a more general question :
What are the function $f:mathbb R rightarrow mathbb R$ such that for all
$sum a_n$ CV (resp. ACV), $sum f(a_n)$ is CV (resp. ACV).
It is a (difficult) exercise to show that for $f:mathbb Rrightarrow mathbb R$,
$$sum a_n ~CV Rightarrow sum f(a_n) ~CV quad textiff quad exists eta>0,exists lambdain mathbb R,forall xin ]-eta,eta[, quad f(x)=lambda x $$
$$sum a_n ~CV Rightarrow sum f(a_n) ~ACV quad textiff quad exists eta>0,forall xin ]-eta,eta[, quad f(x)=0 quadquadquad~~$$
$$sum a_n ~ACV Rightarrow sum f(a_n) ~ACV quad textiff quadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquad$$ $$quad f(0)=0 text and exists eta>0,exists M >0,forall xin ]-eta,eta[, |f(x)|leq M |x| $$
$$sum a_n ACV Rightarrow sum f(a_n) ~CV quad textiff quad sum a_n ~ACV Rightarrow sum f(a_n) ~ACV~~~~~~~~~~~~~~ $$
answered 19 mins ago
Can I play with Mathness
1443
New contributor
Can I play with Mathness is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
1
$sum_n a_n^2$ converges for this one.
– Robert Israel
17 mins ago
oups I answer too fast ^^
– Can I play with Mathness
14 mins ago
Can you correct it? Want to upvote the distinction between ACV and CV but can't bring myself to do it with the error
– Oscar Lanzi
5 mins ago
corrected, I also added a more general statement.
– Can I play with Mathness
3 mins ago
add a comment |Â
up vote
0
down vote
up vote
0
down vote
The simplest example I have in mind is :
$$sum_n=1^+inftyfrac(-1)^nsqrtn $$
Beware that convergent (CV) and absolutely convergent (ACV) can be very different.
Indeed, if $sum_n=1^+infty a_n $ is ACV then, $sum_n=1^+infty a_n^2$ is also ACV. To prove it, you can simply notice that, since $lim a_n=0$ then for $n$ big enough, $|a_n^2|<|a_n|$.
You can also ask yourself a more general question :
What are the function $f:mathbb R rightarrow mathbb R$ such that for all
$sum a_n$ CV (resp. ACV), $sum f(a_n)$ is CV (resp. ACV).
It is a (difficult) exercise to show that for $f:mathbb Rrightarrow mathbb R$,
$$sum a_n ~CV Rightarrow sum f(a_n) ~CV quad textiff quad exists eta>0,exists lambdain mathbb R,forall xin ]-eta,eta[, quad f(x)=lambda x $$
$$sum a_n ~CV Rightarrow sum f(a_n) ~ACV quad textiff quad exists eta>0,forall xin ]-eta,eta[, quad f(x)=0 quadquadquad~~$$
$$sum a_n ~ACV Rightarrow sum f(a_n) ~ACV quad textiff quadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquad$$ $$quad f(0)=0 text and exists eta>0,exists M >0,forall xin ]-eta,eta[, |f(x)|leq M |x| $$
$$sum a_n ACV Rightarrow sum f(a_n) ~CV quad textiff quad sum a_n ~ACV Rightarrow sum f(a_n) ~ACV~~~~~~~~~~~~~~ $$
answered 19 mins ago
Can I play with Mathness
1443
New contributor
Can I play with Mathness is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
The simplest example I have in mind is :
$$sum_n=1^+inftyfrac(-1)^nsqrtn $$
Beware that convergent (CV) and absolutely convergent (ACV) can be very different.
Indeed, if $sum_n=1^+infty a_n $ is ACV then, $sum_n=1^+infty a_n^2$ is also ACV. To prove it, you can simply notice that, since $lim a_n=0$ then for $n$ big enough, $|a_n^2|<|a_n|$.
You can also ask yourself a more general question :
What are the function $f:mathbb R rightarrow mathbb R$ such that for all
$sum a_n$ CV (resp. ACV), $sum f(a_n)$ is CV (resp. ACV).
It is a (difficult) exercise to show that for $f:mathbb Rrightarrow mathbb R$,
$$sum a_n ~CV Rightarrow sum f(a_n) ~CV quad textiff quad exists eta>0,exists lambdain mathbb R,forall xin ]-eta,eta[, quad f(x)=lambda x $$
$$sum a_n ~CV Rightarrow sum f(a_n) ~ACV quad textiff quad exists eta>0,forall xin ]-eta,eta[, quad f(x)=0 quadquadquad~~$$
$$sum a_n ~ACV Rightarrow sum f(a_n) ~ACV quad textiff quadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquad$$ $$quad f(0)=0 text and exists eta>0,exists M >0,forall xin ]-eta,eta[, |f(x)|leq M |x| $$
$$sum a_n ACV Rightarrow sum f(a_n) ~CV quad textiff quad sum a_n ~ACV Rightarrow sum f(a_n) ~ACV~~~~~~~~~~~~~~ $$
answered 19 mins ago
Can I play with Mathness
1443
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edited 4 mins ago
answered 19 mins ago
Can I play with Mathness
1443
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Can I play with Mathness is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
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answered 19 mins ago
Can I play with Mathness
1443
answered 19 mins ago
Can I play with Mathness
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1443
New contributor
Can I play with Mathness is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
New contributor
Can I play with Mathness is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
Can I play with Mathness is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
1
$sum_n a_n^2$ converges for this one.
– Robert Israel
17 mins ago
oups I answer too fast ^^
– Can I play with Mathness
14 mins ago
Can you correct it? Want to upvote the distinction between ACV and CV but can't bring myself to do it with the error
– Oscar Lanzi
5 mins ago
corrected, I also added a more general statement.
– Can I play with Mathness
3 mins ago
add a comment |Â
1
$sum_n a_n^2$ converges for this one.
– Robert Israel
17 mins ago
oups I answer too fast ^^
– Can I play with Mathness
14 mins ago
Can you correct it? Want to upvote the distinction between ACV and CV but can't bring myself to do it with the error
– Oscar Lanzi
5 mins ago
corrected, I also added a more general statement.
– Can I play with Mathness
3 mins ago
1
1
$sum_n a_n^2$ converges for this one.
– Robert Israel
17 mins ago
$sum_n a_n^2$ converges for this one.
– Robert Israel
17 mins ago
oups I answer too fast ^^
– Can I play with Mathness
14 mins ago
oups I answer too fast ^^
– Can I play with Mathness
14 mins ago
Can you correct it? Want to upvote the distinction between ACV and CV but can't bring myself to do it with the error
– Oscar Lanzi
5 mins ago
Can you correct it? Want to upvote the distinction between ACV and CV but can't bring myself to do it with the error
– Oscar Lanzi
5 mins ago
corrected, I also added a more general statement.
– Can I play with Mathness
3 mins ago
corrected, I also added a more general statement.
– Can I play with Mathness
3 mins ago
add a comment |Â
Snop D. is a new contributor. Be nice, and check out our Code of Conduct.
Snop D. is a new contributor. Be nice, and check out our Code of Conduct.
Snop D. is a new contributor. Be nice, and check out our Code of Conduct.
Snop D. is a new contributor. Be nice, and check out our Code of Conduct.
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The implication you said after "because" is true but it's still possible that $sum_n a_n$ converges but not absolutely and $sum_n a_n^2$ does not converge (absolutely nor by any means).
– Saucy O'Path
19 mins ago
@max_zorn That's if $sum a_n$ converges absolutely.
– Robert Israel
15 mins ago
Yes, you are right Robert, sorry!
– max_zorn
12 mins ago