sum of alternating $|f|$ and $|f|^2$
Does anyone know an example of a function $f$ for which the relation
$$
sum_{n=1}^infty (-1)^n |f(n)| < infty \
Longleftrightarrow \
sum_{n=1}^infty (-1)^n |f(n)|^2 < infty
$$
is violated?
Even though the counterexamples are somewhat valid I was more thinking about a differentiable function $f(n)$, possibly even analytic.
In a similar manner: Does the following hold
$$
sum_{n=1}^infty (-1)^n |f(n)|^2 < infty \
Longrightarrow quad sum_{n=1}^infty frac{(-1)^n|f(n)|^2}{1+a^2 |f(n)|^2} < infty
$$
where $a>0$ ? The last one is interesting, because for $a=0$ this matches the assumption and for $a rightarrow infty$ the sum is bounded as well since $left|sum_{n=1}^infty (-1)^nright| leq 1$.
If it is even possible to show
$$
sum_{n=1}^infty frac{(-1)^n|f(n)|^2}{1+a^2 |f(n)|^2} sim {cal O}left(a^{-1-epsilon}right) qquad {rm as} qquad arightarrow infty
$$
and $epsilon>0$ then the integral
$$
frac{1}{pi} int_{-infty}^{infty} sum_{n=1}^infty frac{(-1)^n|f(n)|^2}{1+a^2 |f(n)|^2} , {rm d}a = sum_{n=1}^infty (-1)^n |f(n)| < infty
$$
is well defined and reproduces the first relation.
sequences-and-series summation
asked Dec 1 at 11:50
Diger
1,5941413
add a comment |
Does anyone know an example of a function $f$ for which the relation
$$
sum_{n=1}^infty (-1)^n |f(n)| < infty \
Longleftrightarrow \
sum_{n=1}^infty (-1)^n |f(n)|^2 < infty
$$
is violated?
Even though the counterexamples are somewhat valid I was more thinking about a differentiable function $f(n)$, possibly even analytic.
In a similar manner: Does the following hold
$$
sum_{n=1}^infty (-1)^n |f(n)|^2 < infty \
Longrightarrow quad sum_{n=1}^infty frac{(-1)^n|f(n)|^2}{1+a^2 |f(n)|^2} < infty
$$
where $a>0$ ? The last one is interesting, because for $a=0$ this matches the assumption and for $a rightarrow infty$ the sum is bounded as well since $left|sum_{n=1}^infty (-1)^nright| leq 1$.
If it is even possible to show
$$
sum_{n=1}^infty frac{(-1)^n|f(n)|^2}{1+a^2 |f(n)|^2} sim {cal O}left(a^{-1-epsilon}right) qquad {rm as} qquad arightarrow infty
$$
and $epsilon>0$ then the integral
$$
frac{1}{pi} int_{-infty}^{infty} sum_{n=1}^infty frac{(-1)^n|f(n)|^2}{1+a^2 |f(n)|^2} , {rm d}a = sum_{n=1}^infty (-1)^n |f(n)| < infty
$$
is well defined and reproduces the first relation.
sequences-and-series summation
asked Dec 1 at 11:50
Diger
1,5941413
I think you mean an example for which the relation is violated. a counterexample to violating the biconditional would be an example showing it
– mathworker21
Dec 1 at 12:09
True, I was thinking about a counterexample and ended the sentence wrong.
– Diger
Dec 1 at 12:11
See also math.stackexchange.com/questions/2570953/… The counterexample by I.Browne can be adapted to your case by reindexing with $f(n)=0$ for unwanted terms. This time the first series is convergent while the second is not. In the other answers you get here, it is the opposite.
– zwim
Dec 1 at 13:26
Good example for the other way around, but unfortunately still not analytic.
– Diger
Dec 1 at 13:34
add a comment |
Does anyone know an example of a function $f$ for which the relation
$$
sum_{n=1}^infty (-1)^n |f(n)| < infty \
Longleftrightarrow \
sum_{n=1}^infty (-1)^n |f(n)|^2 < infty
$$
is violated?
Even though the counterexamples are somewhat valid I was more thinking about a differentiable function $f(n)$, possibly even analytic.
In a similar manner: Does the following hold
$$
sum_{n=1}^infty (-1)^n |f(n)|^2 < infty \
Longrightarrow quad sum_{n=1}^infty frac{(-1)^n|f(n)|^2}{1+a^2 |f(n)|^2} < infty
$$
where $a>0$ ? The last one is interesting, because for $a=0$ this matches the assumption and for $a rightarrow infty$ the sum is bounded as well since $left|sum_{n=1}^infty (-1)^nright| leq 1$.
If it is even possible to show
$$
sum_{n=1}^infty frac{(-1)^n|f(n)|^2}{1+a^2 |f(n)|^2} sim {cal O}left(a^{-1-epsilon}right) qquad {rm as} qquad arightarrow infty
$$
and $epsilon>0$ then the integral
$$
frac{1}{pi} int_{-infty}^{infty} sum_{n=1}^infty frac{(-1)^n|f(n)|^2}{1+a^2 |f(n)|^2} , {rm d}a = sum_{n=1}^infty (-1)^n |f(n)| < infty
$$
is well defined and reproduces the first relation.
sequences-and-series summation
asked Dec 1 at 11:50
Diger
1,5941413
Does anyone know an example of a function $f$ for which the relation
$$
sum_{n=1}^infty (-1)^n |f(n)| < infty \
Longleftrightarrow \
sum_{n=1}^infty (-1)^n |f(n)|^2 < infty
$$
is violated?
Even though the counterexamples are somewhat valid I was more thinking about a differentiable function $f(n)$, possibly even analytic.
In a similar manner: Does the following hold
$$
sum_{n=1}^infty (-1)^n |f(n)|^2 < infty \
Longrightarrow quad sum_{n=1}^infty frac{(-1)^n|f(n)|^2}{1+a^2 |f(n)|^2} < infty
$$
where $a>0$ ? The last one is interesting, because for $a=0$ this matches the assumption and for $a rightarrow infty$ the sum is bounded as well since $left|sum_{n=1}^infty (-1)^nright| leq 1$.
If it is even possible to show
$$
sum_{n=1}^infty frac{(-1)^n|f(n)|^2}{1+a^2 |f(n)|^2} sim {cal O}left(a^{-1-epsilon}right) qquad {rm as} qquad arightarrow infty
$$
and $epsilon>0$ then the integral
$$
frac{1}{pi} int_{-infty}^{infty} sum_{n=1}^infty frac{(-1)^n|f(n)|^2}{1+a^2 |f(n)|^2} , {rm d}a = sum_{n=1}^infty (-1)^n |f(n)| < infty
$$
is well defined and reproduces the first relation.
sequences-and-series summation
sequences-and-series summation
asked Dec 1 at 11:50
Diger
1,5941413
asked Dec 1 at 11:50
Diger
1,5941413
edited Dec 2 at 2:45
asked Dec 1 at 11:50
Diger
1,5941413
asked Dec 1 at 11:50
Diger
1,5941413
asked Dec 1 at 11:50
Diger
1,5941413
1,5941413
I think you mean an example for which the relation is violated. a counterexample to violating the biconditional would be an example showing it
– mathworker21
Dec 1 at 12:09
True, I was thinking about a counterexample and ended the sentence wrong.
– Diger
Dec 1 at 12:11
See also math.stackexchange.com/questions/2570953/… The counterexample by I.Browne can be adapted to your case by reindexing with $f(n)=0$ for unwanted terms. This time the first series is convergent while the second is not. In the other answers you get here, it is the opposite.
– zwim
Dec 1 at 13:26
Good example for the other way around, but unfortunately still not analytic.
– Diger
Dec 1 at 13:34
add a comment |
I think you mean an example for which the relation is violated. a counterexample to violating the biconditional would be an example showing it
– mathworker21
Dec 1 at 12:09
True, I was thinking about a counterexample and ended the sentence wrong.
– Diger
Dec 1 at 12:11
See also math.stackexchange.com/questions/2570953/… The counterexample by I.Browne can be adapted to your case by reindexing with $f(n)=0$ for unwanted terms. This time the first series is convergent while the second is not. In the other answers you get here, it is the opposite.
– zwim
Dec 1 at 13:26
Good example for the other way around, but unfortunately still not analytic.
– Diger
Dec 1 at 13:34
I think you mean an example for which the relation is violated. a counterexample to violating the biconditional would be an example showing it
– mathworker21
Dec 1 at 12:09
I think you mean an example for which the relation is violated. a counterexample to violating the biconditional would be an example showing it
– mathworker21
Dec 1 at 12:09
True, I was thinking about a counterexample and ended the sentence wrong.
– Diger
Dec 1 at 12:11
True, I was thinking about a counterexample and ended the sentence wrong.
– Diger
Dec 1 at 12:11
See also math.stackexchange.com/questions/2570953/… The counterexample by I.Browne can be adapted to your case by reindexing with $f(n)=0$ for unwanted terms. This time the first series is convergent while the second is not. In the other answers you get here, it is the opposite.
– zwim
Dec 1 at 13:26
See also math.stackexchange.com/questions/2570953/… The counterexample by I.Browne can be adapted to your case by reindexing with $f(n)=0$ for unwanted terms. This time the first series is convergent while the second is not. In the other answers you get here, it is the opposite.
– zwim
Dec 1 at 13:26
Good example for the other way around, but unfortunately still not analytic.
– Diger
Dec 1 at 13:34
Good example for the other way around, but unfortunately still not analytic.
– Diger
Dec 1 at 13:34
add a comment |
1 Answer
1
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You can put all the weight of the harmonic series on the terms with one sign. For instance
$$
f (n)=begin {cases}1/n,&n text {odd},\ 0,&n text {even}end {cases}
$$
makes the first series divergent and the second convergent.
answered Dec 1 at 12:06
Martin Argerami
123k1176174
add a comment |
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You can put all the weight of the harmonic series on the terms with one sign. For instance
$$
f (n)=begin {cases}1/n,&n text {odd},\ 0,&n text {even}end {cases}
$$
makes the first series divergent and the second convergent.
answered Dec 1 at 12:06
Martin Argerami
123k1176174
add a comment |
You can put all the weight of the harmonic series on the terms with one sign. For instance
$$
f (n)=begin {cases}1/n,&n text {odd},\ 0,&n text {even}end {cases}
$$
makes the first series divergent and the second convergent.
answered Dec 1 at 12:06
Martin Argerami
123k1176174
add a comment |
You can put all the weight of the harmonic series on the terms with one sign. For instance
$$
f (n)=begin {cases}1/n,&n text {odd},\ 0,&n text {even}end {cases}
$$
makes the first series divergent and the second convergent.
answered Dec 1 at 12:06
Martin Argerami
123k1176174
You can put all the weight of the harmonic series on the terms with one sign. For instance
$$
f (n)=begin {cases}1/n,&n text {odd},\ 0,&n text {even}end {cases}
$$
makes the first series divergent and the second convergent.
answered Dec 1 at 12:06
Martin Argerami
123k1176174
answered Dec 1 at 12:06
Martin Argerami
123k1176174
answered Dec 1 at 12:06
Martin Argerami
123k1176174
answered Dec 1 at 12:06
Martin Argerami
123k1176174
123k1176174
add a comment |
add a comment |
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I think you mean an example for which the relation is violated. a counterexample to violating the biconditional would be an example showing it
– mathworker21
Dec 1 at 12:09
True, I was thinking about a counterexample and ended the sentence wrong.
– Diger
Dec 1 at 12:11
See also math.stackexchange.com/questions/2570953/… The counterexample by I.Browne can be adapted to your case by reindexing with $f(n)=0$ for unwanted terms. This time the first series is convergent while the second is not. In the other answers you get here, it is the opposite.
– zwim
Dec 1 at 13:26
Good example for the other way around, but unfortunately still not analytic.
– Diger
Dec 1 at 13:34