Determine the largest real number in [0,2] $beta$ a double series converges.












1












$begingroup$


I wish to determine the largest real number $beta in [0,2]$ for which the following converges:
begin{align}
sum_{j=1}^{infty}sum_{i=1}^{infty} frac{(i^2+j^2)^{beta-1}}{(ij)^3}
end{align}
now, my first attempt is either to try to say, ok, $beta=2$ and then see that $beta$ is to large... but I need to either show that it really is $beta-epsilon$ for where $epsilon$ is arbitrarly small or determine the exact value. So, my second attempt is to use bionomial expansions, where $()_k$ is the falling factorial:



begin{align}
sum_{j=1}^{infty}sum_{i=1}^{infty} frac{1}{i^3j^3} sum_{k=0}^{infty} frac{(beta-1)_k}{k!}(i^2)^{beta-1-k}(j^2)^k
end{align}
and then try to move stuff in...but there i get stuck. The third attempt is to see use Cauchy's condensation criterion, but no dice. any suggestions?






sequences-and-series convergence







share|cite|improve this question









$endgroup$












  • $begingroup$
    To assure the sum is finite, we need to know that we sum finite numbers. And this can be only when $beta<2$.
    $endgroup$
    – Jakobian
    Jan 7 at 17:49












  • $begingroup$
    @Jakobian: No, when $beta=1$ the exponent in the numerator is $0$, so the numerator is $1$ and we get a finite sum. I missed the $-1$ initially, too.
    $endgroup$
    – Ross Millikan
    Jan 7 at 17:50










  • $begingroup$
    @RossMillikan I meant $beta = 2$, not $1$
    $endgroup$
    – Jakobian
    Jan 7 at 17:52


















1












$begingroup$


I wish to determine the largest real number $beta in [0,2]$ for which the following converges:
begin{align}
sum_{j=1}^{infty}sum_{i=1}^{infty} frac{(i^2+j^2)^{beta-1}}{(ij)^3}
end{align}
now, my first attempt is either to try to say, ok, $beta=2$ and then see that $beta$ is to large... but I need to either show that it really is $beta-epsilon$ for where $epsilon$ is arbitrarly small or determine the exact value. So, my second attempt is to use bionomial expansions, where $()_k$ is the falling factorial:



begin{align}
sum_{j=1}^{infty}sum_{i=1}^{infty} frac{1}{i^3j^3} sum_{k=0}^{infty} frac{(beta-1)_k}{k!}(i^2)^{beta-1-k}(j^2)^k
end{align}
and then try to move stuff in...but there i get stuck. The third attempt is to see use Cauchy's condensation criterion, but no dice. any suggestions?






sequences-and-series convergence







share|cite|improve this question









$endgroup$












  • $begingroup$
    To assure the sum is finite, we need to know that we sum finite numbers. And this can be only when $beta<2$.
    $endgroup$
    – Jakobian
    Jan 7 at 17:49












  • $begingroup$
    @Jakobian: No, when $beta=1$ the exponent in the numerator is $0$, so the numerator is $1$ and we get a finite sum. I missed the $-1$ initially, too.
    $endgroup$
    – Ross Millikan
    Jan 7 at 17:50










  • $begingroup$
    @RossMillikan I meant $beta = 2$, not $1$
    $endgroup$
    – Jakobian
    Jan 7 at 17:52
















1












1








1





$begingroup$


I wish to determine the largest real number $beta in [0,2]$ for which the following converges:
begin{align}
sum_{j=1}^{infty}sum_{i=1}^{infty} frac{(i^2+j^2)^{beta-1}}{(ij)^3}
end{align}
now, my first attempt is either to try to say, ok, $beta=2$ and then see that $beta$ is to large... but I need to either show that it really is $beta-epsilon$ for where $epsilon$ is arbitrarly small or determine the exact value. So, my second attempt is to use bionomial expansions, where $()_k$ is the falling factorial:



begin{align}
sum_{j=1}^{infty}sum_{i=1}^{infty} frac{1}{i^3j^3} sum_{k=0}^{infty} frac{(beta-1)_k}{k!}(i^2)^{beta-1-k}(j^2)^k
end{align}
and then try to move stuff in...but there i get stuck. The third attempt is to see use Cauchy's condensation criterion, but no dice. any suggestions?






sequences-and-series convergence







share|cite|improve this question









$endgroup$




I wish to determine the largest real number $beta in [0,2]$ for which the following converges:
begin{align}
sum_{j=1}^{infty}sum_{i=1}^{infty} frac{(i^2+j^2)^{beta-1}}{(ij)^3}
end{align}
now, my first attempt is either to try to say, ok, $beta=2$ and then see that $beta$ is to large... but I need to either show that it really is $beta-epsilon$ for where $epsilon$ is arbitrarly small or determine the exact value. So, my second attempt is to use bionomial expansions, where $()_k$ is the falling factorial:



begin{align}
sum_{j=1}^{infty}sum_{i=1}^{infty} frac{1}{i^3j^3} sum_{k=0}^{infty} frac{(beta-1)_k}{k!}(i^2)^{beta-1-k}(j^2)^k
end{align}
and then try to move stuff in...but there i get stuck. The third attempt is to see use Cauchy's condensation criterion, but no dice. any suggestions?






sequences-and-series convergence




sequences-and-series convergence






share|cite|improve this question













share|cite|improve this question











share|cite|improve this question




share|cite|improve this question










asked Jan 7 at 17:30









edoedo

239




239












  • $begingroup$
    To assure the sum is finite, we need to know that we sum finite numbers. And this can be only when $beta<2$.
    $endgroup$
    – Jakobian
    Jan 7 at 17:49












  • $begingroup$
    @Jakobian: No, when $beta=1$ the exponent in the numerator is $0$, so the numerator is $1$ and we get a finite sum. I missed the $-1$ initially, too.
    $endgroup$
    – Ross Millikan
    Jan 7 at 17:50










  • $begingroup$
    @RossMillikan I meant $beta = 2$, not $1$
    $endgroup$
    – Jakobian
    Jan 7 at 17:52




















  • $begingroup$
    To assure the sum is finite, we need to know that we sum finite numbers. And this can be only when $beta<2$.
    $endgroup$
    – Jakobian
    Jan 7 at 17:49












  • $begingroup$
    @Jakobian: No, when $beta=1$ the exponent in the numerator is $0$, so the numerator is $1$ and we get a finite sum. I missed the $-1$ initially, too.
    $endgroup$
    – Ross Millikan
    Jan 7 at 17:50










  • $begingroup$
    @RossMillikan I meant $beta = 2$, not $1$
    $endgroup$
    – Jakobian
    Jan 7 at 17:52


















$begingroup$
To assure the sum is finite, we need to know that we sum finite numbers. And this can be only when $beta<2$.
$endgroup$
– Jakobian
Jan 7 at 17:49






$begingroup$
To assure the sum is finite, we need to know that we sum finite numbers. And this can be only when $beta<2$.
$endgroup$
– Jakobian
Jan 7 at 17:49














$begingroup$
@Jakobian: No, when $beta=1$ the exponent in the numerator is $0$, so the numerator is $1$ and we get a finite sum. I missed the $-1$ initially, too.
$endgroup$
– Ross Millikan
Jan 7 at 17:50




$begingroup$
@Jakobian: No, when $beta=1$ the exponent in the numerator is $0$, so the numerator is $1$ and we get a finite sum. I missed the $-1$ initially, too.
$endgroup$
– Ross Millikan
Jan 7 at 17:50












$begingroup$
@RossMillikan I meant $beta = 2$, not $1$
$endgroup$
– Jakobian
Jan 7 at 17:52






$begingroup$
@RossMillikan I meant $beta = 2$, not $1$
$endgroup$
– Jakobian
Jan 7 at 17:52












1 Answer
1






active

oldest

votes


















3












$begingroup$

Hints:




  1. Use $$(i^2+j^2)^{beta-1} geq i^{2(beta-1)}$$ to show that $$sum_{i=1}^{infty} frac{(i^2+j^2)^{beta-1}}{(ij)^3} = infty quad text{for $beta=2$}.$$ Hence, $$sum_{j=1}^{infty} sum_{i=1}^{infty} frac{(i^2+j^2)^{beta-1}}{(ij)^3} = infty quad text{for $beta=2$}.$$

  2. Prove (or recall) that $(x+y)^p leq x^p + y^p$ for any $x,y>0$ and $p in (0,1)$.

  3. Show that $$(i^2+j^2)^{beta-1} leq begin{cases} i^{2(beta-1)} + j^{2(beta-1)}, & beta in (1,2) \ 1, & beta in [0,1] end{cases}$$ for any $i,j geq 1$.

  4. Use Step 3 to prove that $$sum_{j=1}^{infty} sum_{i=1}^{infty} frac{(i^2+j^2)^{beta-1}}{(ij)^3} < infty quad text{for $beta in (0,2)$}.$$






share|cite|improve this answer









$endgroup$













  • $begingroup$
    @edo Well, you didn't mention that you already know all of that. It's hard to say what your advisor wants to hear from you; certainly $$sup{beta; text{the series converges for $beta$}}=2$$ but the series fails to converge for $beta=2$. There is really not more to say about this.
    $endgroup$
    – saz
    Jan 7 at 18:21











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1 Answer
1






active

oldest

votes








1 Answer
1






active

oldest

votes









active

oldest

votes






active

oldest

votes









3












$begingroup$

Hints:




  1. Use $$(i^2+j^2)^{beta-1} geq i^{2(beta-1)}$$ to show that $$sum_{i=1}^{infty} frac{(i^2+j^2)^{beta-1}}{(ij)^3} = infty quad text{for $beta=2$}.$$ Hence, $$sum_{j=1}^{infty} sum_{i=1}^{infty} frac{(i^2+j^2)^{beta-1}}{(ij)^3} = infty quad text{for $beta=2$}.$$

  2. Prove (or recall) that $(x+y)^p leq x^p + y^p$ for any $x,y>0$ and $p in (0,1)$.

  3. Show that $$(i^2+j^2)^{beta-1} leq begin{cases} i^{2(beta-1)} + j^{2(beta-1)}, & beta in (1,2) \ 1, & beta in [0,1] end{cases}$$ for any $i,j geq 1$.

  4. Use Step 3 to prove that $$sum_{j=1}^{infty} sum_{i=1}^{infty} frac{(i^2+j^2)^{beta-1}}{(ij)^3} < infty quad text{for $beta in (0,2)$}.$$






share|cite|improve this answer









$endgroup$













  • $begingroup$
    @edo Well, you didn't mention that you already know all of that. It's hard to say what your advisor wants to hear from you; certainly $$sup{beta; text{the series converges for $beta$}}=2$$ but the series fails to converge for $beta=2$. There is really not more to say about this.
    $endgroup$
    – saz
    Jan 7 at 18:21
















3












$begingroup$

Hints:




  1. Use $$(i^2+j^2)^{beta-1} geq i^{2(beta-1)}$$ to show that $$sum_{i=1}^{infty} frac{(i^2+j^2)^{beta-1}}{(ij)^3} = infty quad text{for $beta=2$}.$$ Hence, $$sum_{j=1}^{infty} sum_{i=1}^{infty} frac{(i^2+j^2)^{beta-1}}{(ij)^3} = infty quad text{for $beta=2$}.$$

  2. Prove (or recall) that $(x+y)^p leq x^p + y^p$ for any $x,y>0$ and $p in (0,1)$.

  3. Show that $$(i^2+j^2)^{beta-1} leq begin{cases} i^{2(beta-1)} + j^{2(beta-1)}, & beta in (1,2) \ 1, & beta in [0,1] end{cases}$$ for any $i,j geq 1$.

  4. Use Step 3 to prove that $$sum_{j=1}^{infty} sum_{i=1}^{infty} frac{(i^2+j^2)^{beta-1}}{(ij)^3} < infty quad text{for $beta in (0,2)$}.$$






share|cite|improve this answer









$endgroup$













  • $begingroup$
    @edo Well, you didn't mention that you already know all of that. It's hard to say what your advisor wants to hear from you; certainly $$sup{beta; text{the series converges for $beta$}}=2$$ but the series fails to converge for $beta=2$. There is really not more to say about this.
    $endgroup$
    – saz
    Jan 7 at 18:21














3












3








3





$begingroup$

Hints:




  1. Use $$(i^2+j^2)^{beta-1} geq i^{2(beta-1)}$$ to show that $$sum_{i=1}^{infty} frac{(i^2+j^2)^{beta-1}}{(ij)^3} = infty quad text{for $beta=2$}.$$ Hence, $$sum_{j=1}^{infty} sum_{i=1}^{infty} frac{(i^2+j^2)^{beta-1}}{(ij)^3} = infty quad text{for $beta=2$}.$$

  2. Prove (or recall) that $(x+y)^p leq x^p + y^p$ for any $x,y>0$ and $p in (0,1)$.

  3. Show that $$(i^2+j^2)^{beta-1} leq begin{cases} i^{2(beta-1)} + j^{2(beta-1)}, & beta in (1,2) \ 1, & beta in [0,1] end{cases}$$ for any $i,j geq 1$.

  4. Use Step 3 to prove that $$sum_{j=1}^{infty} sum_{i=1}^{infty} frac{(i^2+j^2)^{beta-1}}{(ij)^3} < infty quad text{for $beta in (0,2)$}.$$






share|cite|improve this answer









$endgroup$



Hints:




  1. Use $$(i^2+j^2)^{beta-1} geq i^{2(beta-1)}$$ to show that $$sum_{i=1}^{infty} frac{(i^2+j^2)^{beta-1}}{(ij)^3} = infty quad text{for $beta=2$}.$$ Hence, $$sum_{j=1}^{infty} sum_{i=1}^{infty} frac{(i^2+j^2)^{beta-1}}{(ij)^3} = infty quad text{for $beta=2$}.$$

  2. Prove (or recall) that $(x+y)^p leq x^p + y^p$ for any $x,y>0$ and $p in (0,1)$.

  3. Show that $$(i^2+j^2)^{beta-1} leq begin{cases} i^{2(beta-1)} + j^{2(beta-1)}, & beta in (1,2) \ 1, & beta in [0,1] end{cases}$$ for any $i,j geq 1$.

  4. Use Step 3 to prove that $$sum_{j=1}^{infty} sum_{i=1}^{infty} frac{(i^2+j^2)^{beta-1}}{(ij)^3} < infty quad text{for $beta in (0,2)$}.$$







share|cite|improve this answer












share|cite|improve this answer



share|cite|improve this answer










answered Jan 7 at 17:50









sazsaz

79.8k860124




79.8k860124












  • $begingroup$
    @edo Well, you didn't mention that you already know all of that. It's hard to say what your advisor wants to hear from you; certainly $$sup{beta; text{the series converges for $beta$}}=2$$ but the series fails to converge for $beta=2$. There is really not more to say about this.
    $endgroup$
    – saz
    Jan 7 at 18:21


















  • $begingroup$
    @edo Well, you didn't mention that you already know all of that. It's hard to say what your advisor wants to hear from you; certainly $$sup{beta; text{the series converges for $beta$}}=2$$ but the series fails to converge for $beta=2$. There is really not more to say about this.
    $endgroup$
    – saz
    Jan 7 at 18:21
















$begingroup$
@edo Well, you didn't mention that you already know all of that. It's hard to say what your advisor wants to hear from you; certainly $$sup{beta; text{the series converges for $beta$}}=2$$ but the series fails to converge for $beta=2$. There is really not more to say about this.
$endgroup$
– saz
Jan 7 at 18:21




$begingroup$
@edo Well, you didn't mention that you already know all of that. It's hard to say what your advisor wants to hear from you; certainly $$sup{beta; text{the series converges for $beta$}}=2$$ but the series fails to converge for $beta=2$. There is really not more to say about this.
$endgroup$
– saz
Jan 7 at 18:21


















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