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Abel-Dini-Type Theorem for convergent series
Contest problem about convergent seriesA generalization of Cauchy's condensation testA convergent series propertyConvergence of the series $sumlimits_ne^-a_n$Improving a proof for a basic property of convergent seriesConvergent series with conditionSize of a convergent series restricted to primesCan the speed of divergence/convergence of a series be controlled arbitrarily?Limit of the series $lim_nrightarrow inftyfrac1s_nsum_k=1^na_kx_k$Absolutely convergent series of complex functions.
$begingroup$
Given are a continuous, strictly increasing and convex function $varphi:[0,infty)to[0,infty)$ and a sequence $(x_j)$ of positive real numbers such that
beginalign*
sum_j=1^inftyvarphi(x_j)
endalign*
converges.
Question: Can we always find a sequence $(lambda _j)$ of positive real numbers which decreases to 0 and so that
beginalign*
sum_j=1^inftyvarphi(x_j/lambda_j)
endalign*
still converges?
This is true for linear $varphi$ by the Abel-Dini-Theorem, and in this case one can take $lambda_j=sqrtx_j+1+x_j+2+ldots$. But what about the general case? Any help is highly appreciated. Thanks in advance!
sequences-and-series convergence
asked Mar 14 at 16:28
sranthropsranthrop
7,1111925
$endgroup$
add a comment |
$begingroup$
Given are a continuous, strictly increasing and convex function $varphi:[0,infty)to[0,infty)$ and a sequence $(x_j)$ of positive real numbers such that
beginalign*
sum_j=1^inftyvarphi(x_j)
endalign*
converges.
Question: Can we always find a sequence $(lambda _j)$ of positive real numbers which decreases to 0 and so that
beginalign*
sum_j=1^inftyvarphi(x_j/lambda_j)
endalign*
still converges?
This is true for linear $varphi$ by the Abel-Dini-Theorem, and in this case one can take $lambda_j=sqrtx_j+1+x_j+2+ldots$. But what about the general case? Any help is highly appreciated. Thanks in advance!
sequences-and-series convergence
asked Mar 14 at 16:28
sranthropsranthrop
7,1111925
$endgroup$
add a comment |
$begingroup$
Given are a continuous, strictly increasing and convex function $varphi:[0,infty)to[0,infty)$ and a sequence $(x_j)$ of positive real numbers such that
beginalign*
sum_j=1^inftyvarphi(x_j)
endalign*
converges.
Question: Can we always find a sequence $(lambda _j)$ of positive real numbers which decreases to 0 and so that
beginalign*
sum_j=1^inftyvarphi(x_j/lambda_j)
endalign*
still converges?
This is true for linear $varphi$ by the Abel-Dini-Theorem, and in this case one can take $lambda_j=sqrtx_j+1+x_j+2+ldots$. But what about the general case? Any help is highly appreciated. Thanks in advance!
sequences-and-series convergence
asked Mar 14 at 16:28
sranthropsranthrop
7,1111925
$endgroup$
Given are a continuous, strictly increasing and convex function $varphi:[0,infty)to[0,infty)$ and a sequence $(x_j)$ of positive real numbers such that
beginalign*
sum_j=1^inftyvarphi(x_j)
endalign*
converges.
Question: Can we always find a sequence $(lambda _j)$ of positive real numbers which decreases to 0 and so that
beginalign*
sum_j=1^inftyvarphi(x_j/lambda_j)
endalign*
still converges?
This is true for linear $varphi$ by the Abel-Dini-Theorem, and in this case one can take $lambda_j=sqrtx_j+1+x_j+2+ldots$. But what about the general case? Any help is highly appreciated. Thanks in advance!
sequences-and-series convergence
sequences-and-series convergence
asked Mar 14 at 16:28
sranthropsranthrop
7,1111925
asked Mar 14 at 16:28
sranthropsranthrop
7,1111925
asked Mar 14 at 16:28
sranthropsranthrop
7,1111925
asked Mar 14 at 16:28
sranthropsranthrop
7,1111925
asked Mar 14 at 16:28
sranthropsranthrop
7,1111925
7,1111925
add a comment |
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
No - here is a counterexample. Define
$$phi(x)=begincases
0&x=0\
exp(-1/x)&0<xleq 1/2\
exp(-2)+4exp(-2)(x-tfrac12)&xgeq 1/2.
endcases$$
For $0<x<1/2$ we have $phi'(x)=x^-2exp(-1/x)$ and $phi''(x)=(x^-4-2x^-3)exp(-1/x)>0$ which ensures that $phi$ is continuous, strictly increasing and convex.
For $0<lambdaleq 1$ and $0<y^lambda<exp(-2)$ we have
$$phi(tfrac1lambda phi^-1(x))=phi(-1/lambdalog(y))=y^lambda.$$
For an extreme example take $x_j=phi^-1(1/j(log j)^2)$ for $jgeq 10$ (and $x_1,dots,x_9=1$ say). Then
$$sum_j=10^infty phi(x_j/lambda)=sum_j=10^infty frac1j^lambda(log j)^2lambda$$
which converges for $lambda=1$ but not for any $lambda<1.$
answered Mar 14 at 18:30
DapDap
18.9k842
$endgroup$
$begingroup$
Great, thank you!
$endgroup$
– sranthrop
Mar 14 at 19:15
add a comment |
Your Answer
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1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
No - here is a counterexample. Define
$$phi(x)=begincases
0&x=0\
exp(-1/x)&0<xleq 1/2\
exp(-2)+4exp(-2)(x-tfrac12)&xgeq 1/2.
endcases$$
For $0<x<1/2$ we have $phi'(x)=x^-2exp(-1/x)$ and $phi''(x)=(x^-4-2x^-3)exp(-1/x)>0$ which ensures that $phi$ is continuous, strictly increasing and convex.
For $0<lambdaleq 1$ and $0<y^lambda<exp(-2)$ we have
$$phi(tfrac1lambda phi^-1(x))=phi(-1/lambdalog(y))=y^lambda.$$
For an extreme example take $x_j=phi^-1(1/j(log j)^2)$ for $jgeq 10$ (and $x_1,dots,x_9=1$ say). Then
$$sum_j=10^infty phi(x_j/lambda)=sum_j=10^infty frac1j^lambda(log j)^2lambda$$
which converges for $lambda=1$ but not for any $lambda<1.$
answered Mar 14 at 18:30
DapDap
18.9k842
$endgroup$
$begingroup$
Great, thank you!
$endgroup$
– sranthrop
Mar 14 at 19:15
add a comment |
$begingroup$
No - here is a counterexample. Define
$$phi(x)=begincases
0&x=0\
exp(-1/x)&0<xleq 1/2\
exp(-2)+4exp(-2)(x-tfrac12)&xgeq 1/2.
endcases$$
For $0<x<1/2$ we have $phi'(x)=x^-2exp(-1/x)$ and $phi''(x)=(x^-4-2x^-3)exp(-1/x)>0$ which ensures that $phi$ is continuous, strictly increasing and convex.
For $0<lambdaleq 1$ and $0<y^lambda<exp(-2)$ we have
$$phi(tfrac1lambda phi^-1(x))=phi(-1/lambdalog(y))=y^lambda.$$
For an extreme example take $x_j=phi^-1(1/j(log j)^2)$ for $jgeq 10$ (and $x_1,dots,x_9=1$ say). Then
$$sum_j=10^infty phi(x_j/lambda)=sum_j=10^infty frac1j^lambda(log j)^2lambda$$
which converges for $lambda=1$ but not for any $lambda<1.$
answered Mar 14 at 18:30
DapDap
18.9k842
$endgroup$
$begingroup$
Great, thank you!
$endgroup$
– sranthrop
Mar 14 at 19:15
add a comment |
$begingroup$
No - here is a counterexample. Define
$$phi(x)=begincases
0&x=0\
exp(-1/x)&0<xleq 1/2\
exp(-2)+4exp(-2)(x-tfrac12)&xgeq 1/2.
endcases$$
For $0<x<1/2$ we have $phi'(x)=x^-2exp(-1/x)$ and $phi''(x)=(x^-4-2x^-3)exp(-1/x)>0$ which ensures that $phi$ is continuous, strictly increasing and convex.
For $0<lambdaleq 1$ and $0<y^lambda<exp(-2)$ we have
$$phi(tfrac1lambda phi^-1(x))=phi(-1/lambdalog(y))=y^lambda.$$
For an extreme example take $x_j=phi^-1(1/j(log j)^2)$ for $jgeq 10$ (and $x_1,dots,x_9=1$ say). Then
$$sum_j=10^infty phi(x_j/lambda)=sum_j=10^infty frac1j^lambda(log j)^2lambda$$
which converges for $lambda=1$ but not for any $lambda<1.$
answered Mar 14 at 18:30
DapDap
18.9k842
$endgroup$
No - here is a counterexample. Define
$$phi(x)=begincases
0&x=0\
exp(-1/x)&0<xleq 1/2\
exp(-2)+4exp(-2)(x-tfrac12)&xgeq 1/2.
endcases$$
For $0<x<1/2$ we have $phi'(x)=x^-2exp(-1/x)$ and $phi''(x)=(x^-4-2x^-3)exp(-1/x)>0$ which ensures that $phi$ is continuous, strictly increasing and convex.
For $0<lambdaleq 1$ and $0<y^lambda<exp(-2)$ we have
$$phi(tfrac1lambda phi^-1(x))=phi(-1/lambdalog(y))=y^lambda.$$
For an extreme example take $x_j=phi^-1(1/j(log j)^2)$ for $jgeq 10$ (and $x_1,dots,x_9=1$ say). Then
$$sum_j=10^infty phi(x_j/lambda)=sum_j=10^infty frac1j^lambda(log j)^2lambda$$
which converges for $lambda=1$ but not for any $lambda<1.$
answered Mar 14 at 18:30
DapDap
18.9k842
answered Mar 14 at 18:30
DapDap
18.9k842
answered Mar 14 at 18:30
DapDap
18.9k842
answered Mar 14 at 18:30
DapDap
18.9k842
18.9k842
$begingroup$
Great, thank you!
$endgroup$
– sranthrop
Mar 14 at 19:15
add a comment |
$begingroup$
Great, thank you!
$endgroup$
– sranthrop
Mar 14 at 19:15
$begingroup$
Great, thank you!
$endgroup$
– sranthrop
Mar 14 at 19:15
$begingroup$
Great, thank you!
$endgroup$
– sranthrop
Mar 14 at 19:15
add a comment |
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