Rate of convergence of a sum of sequence












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$begingroup$


Suppose $a_k=1/k$ and $b_k$ is a positive sequence with $ b_k to 0$. Is there any result that establishes the rate of decay of $b_k$, if $$ sum_k a_kb_k < infty $$



or when can in general $sum_k a_k b_k $ be finite if $b_k to 0$ and is positive for all $k$, and $a_k=1/k$






real-analysis sequences-and-series







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  • $begingroup$
    yes, thanks for pointing out
    $endgroup$
    – user143234
    Dec 24 '18 at 0:58
















0












$begingroup$


Suppose $a_k=1/k$ and $b_k$ is a positive sequence with $ b_k to 0$. Is there any result that establishes the rate of decay of $b_k$, if $$ sum_k a_kb_k < infty $$



or when can in general $sum_k a_k b_k $ be finite if $b_k to 0$ and is positive for all $k$, and $a_k=1/k$






real-analysis sequences-and-series







share|cite|improve this question











$endgroup$












  • $begingroup$
    yes, thanks for pointing out
    $endgroup$
    – user143234
    Dec 24 '18 at 0:58














0












0








0





$begingroup$


Suppose $a_k=1/k$ and $b_k$ is a positive sequence with $ b_k to 0$. Is there any result that establishes the rate of decay of $b_k$, if $$ sum_k a_kb_k < infty $$



or when can in general $sum_k a_k b_k $ be finite if $b_k to 0$ and is positive for all $k$, and $a_k=1/k$






real-analysis sequences-and-series







share|cite|improve this question











$endgroup$




Suppose $a_k=1/k$ and $b_k$ is a positive sequence with $ b_k to 0$. Is there any result that establishes the rate of decay of $b_k$, if $$ sum_k a_kb_k < infty $$



or when can in general $sum_k a_k b_k $ be finite if $b_k to 0$ and is positive for all $k$, and $a_k=1/k$






real-analysis sequences-and-series




real-analysis sequences-and-series






share|cite|improve this question















share|cite|improve this question













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share|cite|improve this question








edited Dec 24 '18 at 0:58







user143234

















asked Dec 24 '18 at 0:53









user143234user143234

1318




1318












  • $begingroup$
    yes, thanks for pointing out
    $endgroup$
    – user143234
    Dec 24 '18 at 0:58


















  • $begingroup$
    yes, thanks for pointing out
    $endgroup$
    – user143234
    Dec 24 '18 at 0:58
















$begingroup$
yes, thanks for pointing out
$endgroup$
– user143234
Dec 24 '18 at 0:58




$begingroup$
yes, thanks for pointing out
$endgroup$
– user143234
Dec 24 '18 at 0:58










1 Answer
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1












$begingroup$

There is no result establishing the rate of decay exactly. This follows from a close analysis of the $p$-series test: that is, $sum n^{-p}$ converges for $p>1$, but the rate of convergence need not be so fast. The sum $sum n^{-1}(log n)^{-p}$ also converges for $p>1$ and its rate of convergence is strictly slower than $sum n^{-p}$.



Applying a similar process, we have $sum (nlog n)^{-1}(loglog n)^{-p}$ converges for $p>1$, so on and so forth. In your notation, this means $b_n=(log n)^{-1}(loglog n)^{-p}$, or to go one step further, $b_n=(log nloglog n)^{-1}(logloglog n)^{-p}$.



In all of these cases, it is true that $b_nllfrac{1}{log n}$, so you at least have a minimum rate of decay. This isn't true in general; see the comments for an example where you can prescribe the rate of decay as slowly as you see fit.



Note: The requirement of positivity for $b_n$ can be loosened to positive for all sufficiently large $n$ since we can just otherwise take $b_n=1$ for the first $N$ terms and then define it using the logarithms for $ngeq N+1$.






share|cite|improve this answer











$endgroup$













  • $begingroup$
    Thanks for the info.
    $endgroup$
    – user143234
    Dec 24 '18 at 1:15










  • $begingroup$
    @user143234: you are welcome.
    $endgroup$
    – Clayton
    Dec 24 '18 at 1:18










  • $begingroup$
    What about say $$b_k = begin{cases}frac{1}{sqrt{log k}} &text{ if } k text{ is a power of } 2\ frac{1}{k^2} &text{ otherwise}end{cases}$$? It does *not* satisfy $b_k ll frac{1}{log k}$...
    $endgroup$
    – Clement C.
    Dec 24 '18 at 1:35












  • $begingroup$
    @ClementC.: Of course. While different from the example I had thought of, your example fits in just fine (that is, I had thought of an example that doesn't satisfy the decay property listed; it doesn't change that the $b_n$ I've defined do satisfy it, though). It seemed like the OP simply wanted to know that there was no definitive, exact rate of decay (for which there isn't).
    $endgroup$
    – Clayton
    Dec 24 '18 at 1:45










  • $begingroup$
    @Clayton: my example is basically to say your sentence "but you can provide some estimate on how quickly it decays" is wrong. You cannot... you can have a subsequence of $(b_k)_k$ going to zero arbitrarily slowly.
    $endgroup$
    – Clement C.
    Dec 24 '18 at 1:48











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$begingroup$

There is no result establishing the rate of decay exactly. This follows from a close analysis of the $p$-series test: that is, $sum n^{-p}$ converges for $p>1$, but the rate of convergence need not be so fast. The sum $sum n^{-1}(log n)^{-p}$ also converges for $p>1$ and its rate of convergence is strictly slower than $sum n^{-p}$.



Applying a similar process, we have $sum (nlog n)^{-1}(loglog n)^{-p}$ converges for $p>1$, so on and so forth. In your notation, this means $b_n=(log n)^{-1}(loglog n)^{-p}$, or to go one step further, $b_n=(log nloglog n)^{-1}(logloglog n)^{-p}$.



In all of these cases, it is true that $b_nllfrac{1}{log n}$, so you at least have a minimum rate of decay. This isn't true in general; see the comments for an example where you can prescribe the rate of decay as slowly as you see fit.



Note: The requirement of positivity for $b_n$ can be loosened to positive for all sufficiently large $n$ since we can just otherwise take $b_n=1$ for the first $N$ terms and then define it using the logarithms for $ngeq N+1$.






share|cite|improve this answer











$endgroup$













  • $begingroup$
    Thanks for the info.
    $endgroup$
    – user143234
    Dec 24 '18 at 1:15










  • $begingroup$
    @user143234: you are welcome.
    $endgroup$
    – Clayton
    Dec 24 '18 at 1:18










  • $begingroup$
    What about say $$b_k = begin{cases}frac{1}{sqrt{log k}} &text{ if } k text{ is a power of } 2\ frac{1}{k^2} &text{ otherwise}end{cases}$$? It does *not* satisfy $b_k ll frac{1}{log k}$...
    $endgroup$
    – Clement C.
    Dec 24 '18 at 1:35












  • $begingroup$
    @ClementC.: Of course. While different from the example I had thought of, your example fits in just fine (that is, I had thought of an example that doesn't satisfy the decay property listed; it doesn't change that the $b_n$ I've defined do satisfy it, though). It seemed like the OP simply wanted to know that there was no definitive, exact rate of decay (for which there isn't).
    $endgroup$
    – Clayton
    Dec 24 '18 at 1:45










  • $begingroup$
    @Clayton: my example is basically to say your sentence "but you can provide some estimate on how quickly it decays" is wrong. You cannot... you can have a subsequence of $(b_k)_k$ going to zero arbitrarily slowly.
    $endgroup$
    – Clement C.
    Dec 24 '18 at 1:48
















1












$begingroup$

There is no result establishing the rate of decay exactly. This follows from a close analysis of the $p$-series test: that is, $sum n^{-p}$ converges for $p>1$, but the rate of convergence need not be so fast. The sum $sum n^{-1}(log n)^{-p}$ also converges for $p>1$ and its rate of convergence is strictly slower than $sum n^{-p}$.



Applying a similar process, we have $sum (nlog n)^{-1}(loglog n)^{-p}$ converges for $p>1$, so on and so forth. In your notation, this means $b_n=(log n)^{-1}(loglog n)^{-p}$, or to go one step further, $b_n=(log nloglog n)^{-1}(logloglog n)^{-p}$.



In all of these cases, it is true that $b_nllfrac{1}{log n}$, so you at least have a minimum rate of decay. This isn't true in general; see the comments for an example where you can prescribe the rate of decay as slowly as you see fit.



Note: The requirement of positivity for $b_n$ can be loosened to positive for all sufficiently large $n$ since we can just otherwise take $b_n=1$ for the first $N$ terms and then define it using the logarithms for $ngeq N+1$.






share|cite|improve this answer











$endgroup$













  • $begingroup$
    Thanks for the info.
    $endgroup$
    – user143234
    Dec 24 '18 at 1:15










  • $begingroup$
    @user143234: you are welcome.
    $endgroup$
    – Clayton
    Dec 24 '18 at 1:18










  • $begingroup$
    What about say $$b_k = begin{cases}frac{1}{sqrt{log k}} &text{ if } k text{ is a power of } 2\ frac{1}{k^2} &text{ otherwise}end{cases}$$? It does *not* satisfy $b_k ll frac{1}{log k}$...
    $endgroup$
    – Clement C.
    Dec 24 '18 at 1:35












  • $begingroup$
    @ClementC.: Of course. While different from the example I had thought of, your example fits in just fine (that is, I had thought of an example that doesn't satisfy the decay property listed; it doesn't change that the $b_n$ I've defined do satisfy it, though). It seemed like the OP simply wanted to know that there was no definitive, exact rate of decay (for which there isn't).
    $endgroup$
    – Clayton
    Dec 24 '18 at 1:45










  • $begingroup$
    @Clayton: my example is basically to say your sentence "but you can provide some estimate on how quickly it decays" is wrong. You cannot... you can have a subsequence of $(b_k)_k$ going to zero arbitrarily slowly.
    $endgroup$
    – Clement C.
    Dec 24 '18 at 1:48














1












1








1





$begingroup$

There is no result establishing the rate of decay exactly. This follows from a close analysis of the $p$-series test: that is, $sum n^{-p}$ converges for $p>1$, but the rate of convergence need not be so fast. The sum $sum n^{-1}(log n)^{-p}$ also converges for $p>1$ and its rate of convergence is strictly slower than $sum n^{-p}$.



Applying a similar process, we have $sum (nlog n)^{-1}(loglog n)^{-p}$ converges for $p>1$, so on and so forth. In your notation, this means $b_n=(log n)^{-1}(loglog n)^{-p}$, or to go one step further, $b_n=(log nloglog n)^{-1}(logloglog n)^{-p}$.



In all of these cases, it is true that $b_nllfrac{1}{log n}$, so you at least have a minimum rate of decay. This isn't true in general; see the comments for an example where you can prescribe the rate of decay as slowly as you see fit.



Note: The requirement of positivity for $b_n$ can be loosened to positive for all sufficiently large $n$ since we can just otherwise take $b_n=1$ for the first $N$ terms and then define it using the logarithms for $ngeq N+1$.






share|cite|improve this answer











$endgroup$



There is no result establishing the rate of decay exactly. This follows from a close analysis of the $p$-series test: that is, $sum n^{-p}$ converges for $p>1$, but the rate of convergence need not be so fast. The sum $sum n^{-1}(log n)^{-p}$ also converges for $p>1$ and its rate of convergence is strictly slower than $sum n^{-p}$.



Applying a similar process, we have $sum (nlog n)^{-1}(loglog n)^{-p}$ converges for $p>1$, so on and so forth. In your notation, this means $b_n=(log n)^{-1}(loglog n)^{-p}$, or to go one step further, $b_n=(log nloglog n)^{-1}(logloglog n)^{-p}$.



In all of these cases, it is true that $b_nllfrac{1}{log n}$, so you at least have a minimum rate of decay. This isn't true in general; see the comments for an example where you can prescribe the rate of decay as slowly as you see fit.



Note: The requirement of positivity for $b_n$ can be loosened to positive for all sufficiently large $n$ since we can just otherwise take $b_n=1$ for the first $N$ terms and then define it using the logarithms for $ngeq N+1$.







share|cite|improve this answer














share|cite|improve this answer



share|cite|improve this answer








edited Dec 24 '18 at 1:50

























answered Dec 24 '18 at 1:09









ClaytonClayton

19.2k33286




19.2k33286












  • $begingroup$
    Thanks for the info.
    $endgroup$
    – user143234
    Dec 24 '18 at 1:15










  • $begingroup$
    @user143234: you are welcome.
    $endgroup$
    – Clayton
    Dec 24 '18 at 1:18










  • $begingroup$
    What about say $$b_k = begin{cases}frac{1}{sqrt{log k}} &text{ if } k text{ is a power of } 2\ frac{1}{k^2} &text{ otherwise}end{cases}$$? It does *not* satisfy $b_k ll frac{1}{log k}$...
    $endgroup$
    – Clement C.
    Dec 24 '18 at 1:35












  • $begingroup$
    @ClementC.: Of course. While different from the example I had thought of, your example fits in just fine (that is, I had thought of an example that doesn't satisfy the decay property listed; it doesn't change that the $b_n$ I've defined do satisfy it, though). It seemed like the OP simply wanted to know that there was no definitive, exact rate of decay (for which there isn't).
    $endgroup$
    – Clayton
    Dec 24 '18 at 1:45










  • $begingroup$
    @Clayton: my example is basically to say your sentence "but you can provide some estimate on how quickly it decays" is wrong. You cannot... you can have a subsequence of $(b_k)_k$ going to zero arbitrarily slowly.
    $endgroup$
    – Clement C.
    Dec 24 '18 at 1:48


















  • $begingroup$
    Thanks for the info.
    $endgroup$
    – user143234
    Dec 24 '18 at 1:15










  • $begingroup$
    @user143234: you are welcome.
    $endgroup$
    – Clayton
    Dec 24 '18 at 1:18










  • $begingroup$
    What about say $$b_k = begin{cases}frac{1}{sqrt{log k}} &text{ if } k text{ is a power of } 2\ frac{1}{k^2} &text{ otherwise}end{cases}$$? It does *not* satisfy $b_k ll frac{1}{log k}$...
    $endgroup$
    – Clement C.
    Dec 24 '18 at 1:35












  • $begingroup$
    @ClementC.: Of course. While different from the example I had thought of, your example fits in just fine (that is, I had thought of an example that doesn't satisfy the decay property listed; it doesn't change that the $b_n$ I've defined do satisfy it, though). It seemed like the OP simply wanted to know that there was no definitive, exact rate of decay (for which there isn't).
    $endgroup$
    – Clayton
    Dec 24 '18 at 1:45










  • $begingroup$
    @Clayton: my example is basically to say your sentence "but you can provide some estimate on how quickly it decays" is wrong. You cannot... you can have a subsequence of $(b_k)_k$ going to zero arbitrarily slowly.
    $endgroup$
    – Clement C.
    Dec 24 '18 at 1:48
















$begingroup$
Thanks for the info.
$endgroup$
– user143234
Dec 24 '18 at 1:15




$begingroup$
Thanks for the info.
$endgroup$
– user143234
Dec 24 '18 at 1:15












$begingroup$
@user143234: you are welcome.
$endgroup$
– Clayton
Dec 24 '18 at 1:18




$begingroup$
@user143234: you are welcome.
$endgroup$
– Clayton
Dec 24 '18 at 1:18












$begingroup$
What about say $$b_k = begin{cases}frac{1}{sqrt{log k}} &text{ if } k text{ is a power of } 2\ frac{1}{k^2} &text{ otherwise}end{cases}$$? It does *not* satisfy $b_k ll frac{1}{log k}$...
$endgroup$
– Clement C.
Dec 24 '18 at 1:35






$begingroup$
What about say $$b_k = begin{cases}frac{1}{sqrt{log k}} &text{ if } k text{ is a power of } 2\ frac{1}{k^2} &text{ otherwise}end{cases}$$? It does *not* satisfy $b_k ll frac{1}{log k}$...
$endgroup$
– Clement C.
Dec 24 '18 at 1:35














$begingroup$
@ClementC.: Of course. While different from the example I had thought of, your example fits in just fine (that is, I had thought of an example that doesn't satisfy the decay property listed; it doesn't change that the $b_n$ I've defined do satisfy it, though). It seemed like the OP simply wanted to know that there was no definitive, exact rate of decay (for which there isn't).
$endgroup$
– Clayton
Dec 24 '18 at 1:45




$begingroup$
@ClementC.: Of course. While different from the example I had thought of, your example fits in just fine (that is, I had thought of an example that doesn't satisfy the decay property listed; it doesn't change that the $b_n$ I've defined do satisfy it, though). It seemed like the OP simply wanted to know that there was no definitive, exact rate of decay (for which there isn't).
$endgroup$
– Clayton
Dec 24 '18 at 1:45












$begingroup$
@Clayton: my example is basically to say your sentence "but you can provide some estimate on how quickly it decays" is wrong. You cannot... you can have a subsequence of $(b_k)_k$ going to zero arbitrarily slowly.
$endgroup$
– Clement C.
Dec 24 '18 at 1:48




$begingroup$
@Clayton: my example is basically to say your sentence "but you can provide some estimate on how quickly it decays" is wrong. You cannot... you can have a subsequence of $(b_k)_k$ going to zero arbitrarily slowly.
$endgroup$
– Clement C.
Dec 24 '18 at 1:48


















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