$sum a_{n}$ converges but $sum a_{n}^2 $ diverges?
up vote
2
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 limits convergence
New contributor
add a comment |
up vote
2
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 limits convergence
New contributor
1
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
Nov 4 at 23:05
@max_zorn That's if $sum a_n$ converges absolutely.
– Robert Israel
Nov 4 at 23:09
Yes, you are right Robert, sorry!
– max_zorn
Nov 4 at 23:12
add a comment |
up vote
2
down vote
favorite
up vote
2
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 limits convergence
New contributor
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 limits convergence
real-analysis sequences-and-series limits convergence
New contributor
New contributor
edited Nov 5 at 7:59
user21820
37.8k441147
37.8k441147
New contributor
asked Nov 4 at 23:03
Snop D.
111
111
New contributor
New contributor
1
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
Nov 4 at 23:05
@max_zorn That's if $sum a_n$ converges absolutely.
– Robert Israel
Nov 4 at 23:09
Yes, you are right Robert, sorry!
– max_zorn
Nov 4 at 23:12
add a comment |
1
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
Nov 4 at 23:05
@max_zorn That's if $sum a_n$ converges absolutely.
– Robert Israel
Nov 4 at 23:09
Yes, you are right Robert, sorry!
– max_zorn
Nov 4 at 23:12
1
1
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
Nov 4 at 23:05
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
Nov 4 at 23:05
@max_zorn That's if $sum a_n$ converges absolutely.
– Robert Israel
Nov 4 at 23:09
@max_zorn That's if $sum a_n$ converges absolutely.
– Robert Israel
Nov 4 at 23:09
Yes, you are right Robert, sorry!
– max_zorn
Nov 4 at 23:12
Yes, you are right Robert, sorry!
– max_zorn
Nov 4 at 23:12
add a comment |
4 Answers
4
active
oldest
votes
up vote
12
down vote
The alternating series test gives a wealth of examples. Take
$$
a_n=(-1)^n/sqrt{n}
$$
for example.
You beat me to it by $14$ seconds.
– Robert Israel
Nov 4 at 23:07
5
Are there any that don't involve a $(-1)^n$ term?
– Ingolifs
Nov 5 at 4:27
1
@Ingolifs: $sin$. But if the sequence is non-negative, it is a basic exercise to prove that the implication holds.
– user21820
Nov 5 at 8:00
add a comment |
up vote
10
down vote
Try $$sum_{n=1}^infty frac{(-1)^n}{sqrt{n}}$$
add a comment |
up vote
3
down vote
More generally,
if $a_n = dfrac{(-1)^n}{n^{1/(2m)}}$,
then
$sum_{n=1}^{infty} a_n
$
converges for integer $m ge 0$
and
$sum_{n=1}^{infty} a_n^{2m}
$
diverges.
add a comment |
up vote
0
down vote
The simplest example I have in mind is :
$$sum_{n=1}^{+infty}frac{(-1)^n}{sqrt{n}} $$
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 $f:mathbb Rrightarrow mathbb R$,
$$sum a_n ~CV Rightarrow sum f(a_n) ~CV quad text{iff} quad exists eta>0,exists lambdain mathbb R,forall xin ]-eta,eta[, quad f(x)=lambda x $$
$$sum a_n ~CV Rightarrow sum f(a_n) ~ACV quad text{iff} quad exists eta>0,forall xin ]-eta,eta[, quad f(x)=0 quadquadquad~~$$
$$sum a_n ~ACV Rightarrow sum f(a_n) ~ACV quad text{iff} quadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquad$$ $$quad f(0)=0 text{ and }exists eta>0,exists M >0,forall xin ]-eta,eta[, |f(x)|leq M |x| $$
$$sum a_n ACV Rightarrow sum f(a_n) ~CV quad text{iff} quad sum a_n ~ACV Rightarrow sum f(a_n) ~ACV~~~~~~~~~~~~~~ $$
New contributor
1
$sum_n a_n^2$ converges for this one.
– Robert Israel
Nov 4 at 23:07
oups I answer too fast ^^
– Can I play with Mathness
Nov 4 at 23:10
3
I have a PhD in math and my eyes glaze over when reading this post because of the unnecessary jumble of symbols. Words are generally preferable for longer statements.
– Matt Samuel
Nov 5 at 1:23
2
While I realize it's a standard notation, $xin]-eta,eta[$ just looks so horrible--the unbalanced brackets make it hard to parse. $xin(-eta,eta)$ is much better, though $|x| < eta$ is better than both.
– eyeballfrog
Nov 5 at 2:33
1
I'm in agreement with the comment by Matt above, for me it's almost unreadable (in the sense that it takes way too much effort to parse it). A (IMO) better way of presenting the first statement would be something like "If $sum a_n$ converges then $sum f(a_n)$ converges if $f(x)$ is proportional to $x$ in a neighborhood of $x=0$."
– Winther
Nov 5 at 2:34
|
show 4 more comments
4 Answers
4
active
oldest
votes
4 Answers
4
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
12
down vote
The alternating series test gives a wealth of examples. Take
$$
a_n=(-1)^n/sqrt{n}
$$
for example.
You beat me to it by $14$ seconds.
– Robert Israel
Nov 4 at 23:07
5
Are there any that don't involve a $(-1)^n$ term?
– Ingolifs
Nov 5 at 4:27
1
@Ingolifs: $sin$. But if the sequence is non-negative, it is a basic exercise to prove that the implication holds.
– user21820
Nov 5 at 8:00
add a comment |
up vote
12
down vote
The alternating series test gives a wealth of examples. Take
$$
a_n=(-1)^n/sqrt{n}
$$
for example.
You beat me to it by $14$ seconds.
– Robert Israel
Nov 4 at 23:07
5
Are there any that don't involve a $(-1)^n$ term?
– Ingolifs
Nov 5 at 4:27
1
@Ingolifs: $sin$. But if the sequence is non-negative, it is a basic exercise to prove that the implication holds.
– user21820
Nov 5 at 8:00
add a comment |
up vote
12
down vote
up vote
12
down vote
The alternating series test gives a wealth of examples. Take
$$
a_n=(-1)^n/sqrt{n}
$$
for example.
The alternating series test gives a wealth of examples. Take
$$
a_n=(-1)^n/sqrt{n}
$$
for example.
answered Nov 4 at 23:05
Foobaz John
19.3k41248
19.3k41248
You beat me to it by $14$ seconds.
– Robert Israel
Nov 4 at 23:07
5
Are there any that don't involve a $(-1)^n$ term?
– Ingolifs
Nov 5 at 4:27
1
@Ingolifs: $sin$. But if the sequence is non-negative, it is a basic exercise to prove that the implication holds.
– user21820
Nov 5 at 8:00
add a comment |
You beat me to it by $14$ seconds.
– Robert Israel
Nov 4 at 23:07
5
Are there any that don't involve a $(-1)^n$ term?
– Ingolifs
Nov 5 at 4:27
1
@Ingolifs: $sin$. But if the sequence is non-negative, it is a basic exercise to prove that the implication holds.
– user21820
Nov 5 at 8:00
You beat me to it by $14$ seconds.
– Robert Israel
Nov 4 at 23:07
You beat me to it by $14$ seconds.
– Robert Israel
Nov 4 at 23:07
5
5
Are there any that don't involve a $(-1)^n$ term?
– Ingolifs
Nov 5 at 4:27
Are there any that don't involve a $(-1)^n$ term?
– Ingolifs
Nov 5 at 4:27
1
1
@Ingolifs: $sin$. But if the sequence is non-negative, it is a basic exercise to prove that the implication holds.
– user21820
Nov 5 at 8:00
@Ingolifs: $sin$. But if the sequence is non-negative, it is a basic exercise to prove that the implication holds.
– user21820
Nov 5 at 8:00
add a comment |
up vote
10
down vote
Try $$sum_{n=1}^infty frac{(-1)^n}{sqrt{n}}$$
add a comment |
up vote
10
down vote
Try $$sum_{n=1}^infty frac{(-1)^n}{sqrt{n}}$$
add a comment |
up vote
10
down vote
up vote
10
down vote
Try $$sum_{n=1}^infty frac{(-1)^n}{sqrt{n}}$$
Try $$sum_{n=1}^infty frac{(-1)^n}{sqrt{n}}$$
answered Nov 4 at 23:06
Robert Israel
312k23206452
312k23206452
add a comment |
add a comment |
up vote
3
down vote
More generally,
if $a_n = dfrac{(-1)^n}{n^{1/(2m)}}$,
then
$sum_{n=1}^{infty} a_n
$
converges for integer $m ge 0$
and
$sum_{n=1}^{infty} a_n^{2m}
$
diverges.
add a comment |
up vote
3
down vote
More generally,
if $a_n = dfrac{(-1)^n}{n^{1/(2m)}}$,
then
$sum_{n=1}^{infty} a_n
$
converges for integer $m ge 0$
and
$sum_{n=1}^{infty} a_n^{2m}
$
diverges.
add a comment |
up vote
3
down vote
up vote
3
down vote
More generally,
if $a_n = dfrac{(-1)^n}{n^{1/(2m)}}$,
then
$sum_{n=1}^{infty} a_n
$
converges for integer $m ge 0$
and
$sum_{n=1}^{infty} a_n^{2m}
$
diverges.
More generally,
if $a_n = dfrac{(-1)^n}{n^{1/(2m)}}$,
then
$sum_{n=1}^{infty} a_n
$
converges for integer $m ge 0$
and
$sum_{n=1}^{infty} a_n^{2m}
$
diverges.
answered Nov 4 at 23:29
marty cohen
71k546122
71k546122
add a comment |
add a comment |
up vote
0
down vote
The simplest example I have in mind is :
$$sum_{n=1}^{+infty}frac{(-1)^n}{sqrt{n}} $$
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 $f:mathbb Rrightarrow mathbb R$,
$$sum a_n ~CV Rightarrow sum f(a_n) ~CV quad text{iff} quad exists eta>0,exists lambdain mathbb R,forall xin ]-eta,eta[, quad f(x)=lambda x $$
$$sum a_n ~CV Rightarrow sum f(a_n) ~ACV quad text{iff} quad exists eta>0,forall xin ]-eta,eta[, quad f(x)=0 quadquadquad~~$$
$$sum a_n ~ACV Rightarrow sum f(a_n) ~ACV quad text{iff} quadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquad$$ $$quad f(0)=0 text{ and }exists eta>0,exists M >0,forall xin ]-eta,eta[, |f(x)|leq M |x| $$
$$sum a_n ACV Rightarrow sum f(a_n) ~CV quad text{iff} quad sum a_n ~ACV Rightarrow sum f(a_n) ~ACV~~~~~~~~~~~~~~ $$
New contributor
1
$sum_n a_n^2$ converges for this one.
– Robert Israel
Nov 4 at 23:07
oups I answer too fast ^^
– Can I play with Mathness
Nov 4 at 23:10
3
I have a PhD in math and my eyes glaze over when reading this post because of the unnecessary jumble of symbols. Words are generally preferable for longer statements.
– Matt Samuel
Nov 5 at 1:23
2
While I realize it's a standard notation, $xin]-eta,eta[$ just looks so horrible--the unbalanced brackets make it hard to parse. $xin(-eta,eta)$ is much better, though $|x| < eta$ is better than both.
– eyeballfrog
Nov 5 at 2:33
1
I'm in agreement with the comment by Matt above, for me it's almost unreadable (in the sense that it takes way too much effort to parse it). A (IMO) better way of presenting the first statement would be something like "If $sum a_n$ converges then $sum f(a_n)$ converges if $f(x)$ is proportional to $x$ in a neighborhood of $x=0$."
– Winther
Nov 5 at 2:34
|
show 4 more comments
up vote
0
down vote
The simplest example I have in mind is :
$$sum_{n=1}^{+infty}frac{(-1)^n}{sqrt{n}} $$
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 $f:mathbb Rrightarrow mathbb R$,
$$sum a_n ~CV Rightarrow sum f(a_n) ~CV quad text{iff} quad exists eta>0,exists lambdain mathbb R,forall xin ]-eta,eta[, quad f(x)=lambda x $$
$$sum a_n ~CV Rightarrow sum f(a_n) ~ACV quad text{iff} quad exists eta>0,forall xin ]-eta,eta[, quad f(x)=0 quadquadquad~~$$
$$sum a_n ~ACV Rightarrow sum f(a_n) ~ACV quad text{iff} quadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquad$$ $$quad f(0)=0 text{ and }exists eta>0,exists M >0,forall xin ]-eta,eta[, |f(x)|leq M |x| $$
$$sum a_n ACV Rightarrow sum f(a_n) ~CV quad text{iff} quad sum a_n ~ACV Rightarrow sum f(a_n) ~ACV~~~~~~~~~~~~~~ $$
New contributor
1
$sum_n a_n^2$ converges for this one.
– Robert Israel
Nov 4 at 23:07
oups I answer too fast ^^
– Can I play with Mathness
Nov 4 at 23:10
3
I have a PhD in math and my eyes glaze over when reading this post because of the unnecessary jumble of symbols. Words are generally preferable for longer statements.
– Matt Samuel
Nov 5 at 1:23
2
While I realize it's a standard notation, $xin]-eta,eta[$ just looks so horrible--the unbalanced brackets make it hard to parse. $xin(-eta,eta)$ is much better, though $|x| < eta$ is better than both.
– eyeballfrog
Nov 5 at 2:33
1
I'm in agreement with the comment by Matt above, for me it's almost unreadable (in the sense that it takes way too much effort to parse it). A (IMO) better way of presenting the first statement would be something like "If $sum a_n$ converges then $sum f(a_n)$ converges if $f(x)$ is proportional to $x$ in a neighborhood of $x=0$."
– Winther
Nov 5 at 2:34
|
show 4 more comments
up vote
0
down vote
up vote
0
down vote
The simplest example I have in mind is :
$$sum_{n=1}^{+infty}frac{(-1)^n}{sqrt{n}} $$
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 $f:mathbb Rrightarrow mathbb R$,
$$sum a_n ~CV Rightarrow sum f(a_n) ~CV quad text{iff} quad exists eta>0,exists lambdain mathbb R,forall xin ]-eta,eta[, quad f(x)=lambda x $$
$$sum a_n ~CV Rightarrow sum f(a_n) ~ACV quad text{iff} quad exists eta>0,forall xin ]-eta,eta[, quad f(x)=0 quadquadquad~~$$
$$sum a_n ~ACV Rightarrow sum f(a_n) ~ACV quad text{iff} quadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquad$$ $$quad f(0)=0 text{ and }exists eta>0,exists M >0,forall xin ]-eta,eta[, |f(x)|leq M |x| $$
$$sum a_n ACV Rightarrow sum f(a_n) ~CV quad text{iff} quad sum a_n ~ACV Rightarrow sum f(a_n) ~ACV~~~~~~~~~~~~~~ $$
New contributor
The simplest example I have in mind is :
$$sum_{n=1}^{+infty}frac{(-1)^n}{sqrt{n}} $$
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 $f:mathbb Rrightarrow mathbb R$,
$$sum a_n ~CV Rightarrow sum f(a_n) ~CV quad text{iff} quad exists eta>0,exists lambdain mathbb R,forall xin ]-eta,eta[, quad f(x)=lambda x $$
$$sum a_n ~CV Rightarrow sum f(a_n) ~ACV quad text{iff} quad exists eta>0,forall xin ]-eta,eta[, quad f(x)=0 quadquadquad~~$$
$$sum a_n ~ACV Rightarrow sum f(a_n) ~ACV quad text{iff} quadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquad$$ $$quad f(0)=0 text{ and }exists eta>0,exists M >0,forall xin ]-eta,eta[, |f(x)|leq M |x| $$
$$sum a_n ACV Rightarrow sum f(a_n) ~CV quad text{iff} quad sum a_n ~ACV Rightarrow sum f(a_n) ~ACV~~~~~~~~~~~~~~ $$
New contributor
edited Nov 4 at 23:20
New contributor
answered Nov 4 at 23:05
Can I play with Mathness
3544
3544
New contributor
New contributor
1
$sum_n a_n^2$ converges for this one.
– Robert Israel
Nov 4 at 23:07
oups I answer too fast ^^
– Can I play with Mathness
Nov 4 at 23:10
3
I have a PhD in math and my eyes glaze over when reading this post because of the unnecessary jumble of symbols. Words are generally preferable for longer statements.
– Matt Samuel
Nov 5 at 1:23
2
While I realize it's a standard notation, $xin]-eta,eta[$ just looks so horrible--the unbalanced brackets make it hard to parse. $xin(-eta,eta)$ is much better, though $|x| < eta$ is better than both.
– eyeballfrog
Nov 5 at 2:33
1
I'm in agreement with the comment by Matt above, for me it's almost unreadable (in the sense that it takes way too much effort to parse it). A (IMO) better way of presenting the first statement would be something like "If $sum a_n$ converges then $sum f(a_n)$ converges if $f(x)$ is proportional to $x$ in a neighborhood of $x=0$."
– Winther
Nov 5 at 2:34
|
show 4 more comments
1
$sum_n a_n^2$ converges for this one.
– Robert Israel
Nov 4 at 23:07
oups I answer too fast ^^
– Can I play with Mathness
Nov 4 at 23:10
3
I have a PhD in math and my eyes glaze over when reading this post because of the unnecessary jumble of symbols. Words are generally preferable for longer statements.
– Matt Samuel
Nov 5 at 1:23
2
While I realize it's a standard notation, $xin]-eta,eta[$ just looks so horrible--the unbalanced brackets make it hard to parse. $xin(-eta,eta)$ is much better, though $|x| < eta$ is better than both.
– eyeballfrog
Nov 5 at 2:33
1
I'm in agreement with the comment by Matt above, for me it's almost unreadable (in the sense that it takes way too much effort to parse it). A (IMO) better way of presenting the first statement would be something like "If $sum a_n$ converges then $sum f(a_n)$ converges if $f(x)$ is proportional to $x$ in a neighborhood of $x=0$."
– Winther
Nov 5 at 2:34
1
1
$sum_n a_n^2$ converges for this one.
– Robert Israel
Nov 4 at 23:07
$sum_n a_n^2$ converges for this one.
– Robert Israel
Nov 4 at 23:07
oups I answer too fast ^^
– Can I play with Mathness
Nov 4 at 23:10
oups I answer too fast ^^
– Can I play with Mathness
Nov 4 at 23:10
3
3
I have a PhD in math and my eyes glaze over when reading this post because of the unnecessary jumble of symbols. Words are generally preferable for longer statements.
– Matt Samuel
Nov 5 at 1:23
I have a PhD in math and my eyes glaze over when reading this post because of the unnecessary jumble of symbols. Words are generally preferable for longer statements.
– Matt Samuel
Nov 5 at 1:23
2
2
While I realize it's a standard notation, $xin]-eta,eta[$ just looks so horrible--the unbalanced brackets make it hard to parse. $xin(-eta,eta)$ is much better, though $|x| < eta$ is better than both.
– eyeballfrog
Nov 5 at 2:33
While I realize it's a standard notation, $xin]-eta,eta[$ just looks so horrible--the unbalanced brackets make it hard to parse. $xin(-eta,eta)$ is much better, though $|x| < eta$ is better than both.
– eyeballfrog
Nov 5 at 2:33
1
1
I'm in agreement with the comment by Matt above, for me it's almost unreadable (in the sense that it takes way too much effort to parse it). A (IMO) better way of presenting the first statement would be something like "If $sum a_n$ converges then $sum f(a_n)$ converges if $f(x)$ is proportional to $x$ in a neighborhood of $x=0$."
– Winther
Nov 5 at 2:34
I'm in agreement with the comment by Matt above, for me it's almost unreadable (in the sense that it takes way too much effort to parse it). A (IMO) better way of presenting the first statement would be something like "If $sum a_n$ converges then $sum f(a_n)$ converges if $f(x)$ is proportional to $x$ in a neighborhood of $x=0$."
– Winther
Nov 5 at 2:34
|
show 4 more comments
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
Nov 4 at 23:05
@max_zorn That's if $sum a_n$ converges absolutely.
– Robert Israel
Nov 4 at 23:09
Yes, you are right Robert, sorry!
– max_zorn
Nov 4 at 23:12