Existence of $a_k$ such that $sum_k a_kb_k<infty$ and $sum_k a_k=infty$ given $b_kto 0$
$begingroup$
I was working with a problem from functional analysis. I reduced the problem to the following problem:
Let $b_k>0$ be a decreasing sequence converging to $0$. Does there exist a non-negative sequence $a_k$ such that
$$sum_{kgeq 1} b_ka_k<infty$$
while at the same time
$$sum_{kgeq 1} a_k = infty$$
If I know what $b_k$ looks like, then I guess it is not hard to find out how to choose $a_k$. However we don't know what $b_k$ looks like and neither in which rate it decays. I really don't know what to do except that I have tried something like choosing for $n>m$
begin{align}
sum_{k=m}^n a_k = frac 1 {sqrt {b_mb_n^varepsilon}}
end{align}
for some $varepsilon>0$, so that the sequence is not Cauchy, that leads to the divergence of $sum_k a_k$ which is nice. At the same time I get
begin{align}
sum_{k=m}^n b_ka_k leq b_m sum_{k=m}^na_k =sqrt{frac{b_m}{b_n^varepsilon}}
end{align}
This tells us that, if there is $varepsilon$ such that
$$sqrt{frac{b_m}{b_n^varepsilon}}to 0 text{ as } n,mtoinfty$$
we are done. However, I don't think that will work, since $varepsilon$ is fixed...
Question. How can the problem be solved?
real-analysis sequences-and-series cauchy-sequences
edited Jan 10 at 19:50
David Mitra
63.1k6100164
asked Jan 10 at 19:22
ShashiShashi
7,1731528
$endgroup$
add a comment |
$begingroup$
I was working with a problem from functional analysis. I reduced the problem to the following problem:
Let $b_k>0$ be a decreasing sequence converging to $0$. Does there exist a non-negative sequence $a_k$ such that
$$sum_{kgeq 1} b_ka_k<infty$$
while at the same time
$$sum_{kgeq 1} a_k = infty$$
If I know what $b_k$ looks like, then I guess it is not hard to find out how to choose $a_k$. However we don't know what $b_k$ looks like and neither in which rate it decays. I really don't know what to do except that I have tried something like choosing for $n>m$
begin{align}
sum_{k=m}^n a_k = frac 1 {sqrt {b_mb_n^varepsilon}}
end{align}
for some $varepsilon>0$, so that the sequence is not Cauchy, that leads to the divergence of $sum_k a_k$ which is nice. At the same time I get
begin{align}
sum_{k=m}^n b_ka_k leq b_m sum_{k=m}^na_k =sqrt{frac{b_m}{b_n^varepsilon}}
end{align}
This tells us that, if there is $varepsilon$ such that
$$sqrt{frac{b_m}{b_n^varepsilon}}to 0 text{ as } n,mtoinfty$$
we are done. However, I don't think that will work, since $varepsilon$ is fixed...
Question. How can the problem be solved?
real-analysis sequences-and-series cauchy-sequences
edited Jan 10 at 19:50
David Mitra
63.1k6100164
asked Jan 10 at 19:22
ShashiShashi
7,1731528
$endgroup$
add a comment |
$begingroup$
I was working with a problem from functional analysis. I reduced the problem to the following problem:
Let $b_k>0$ be a decreasing sequence converging to $0$. Does there exist a non-negative sequence $a_k$ such that
$$sum_{kgeq 1} b_ka_k<infty$$
while at the same time
$$sum_{kgeq 1} a_k = infty$$
If I know what $b_k$ looks like, then I guess it is not hard to find out how to choose $a_k$. However we don't know what $b_k$ looks like and neither in which rate it decays. I really don't know what to do except that I have tried something like choosing for $n>m$
begin{align}
sum_{k=m}^n a_k = frac 1 {sqrt {b_mb_n^varepsilon}}
end{align}
for some $varepsilon>0$, so that the sequence is not Cauchy, that leads to the divergence of $sum_k a_k$ which is nice. At the same time I get
begin{align}
sum_{k=m}^n b_ka_k leq b_m sum_{k=m}^na_k =sqrt{frac{b_m}{b_n^varepsilon}}
end{align}
This tells us that, if there is $varepsilon$ such that
$$sqrt{frac{b_m}{b_n^varepsilon}}to 0 text{ as } n,mtoinfty$$
we are done. However, I don't think that will work, since $varepsilon$ is fixed...
Question. How can the problem be solved?
real-analysis sequences-and-series cauchy-sequences
edited Jan 10 at 19:50
David Mitra
63.1k6100164
asked Jan 10 at 19:22
ShashiShashi
7,1731528
$endgroup$
I was working with a problem from functional analysis. I reduced the problem to the following problem:
Let $b_k>0$ be a decreasing sequence converging to $0$. Does there exist a non-negative sequence $a_k$ such that
$$sum_{kgeq 1} b_ka_k<infty$$
while at the same time
$$sum_{kgeq 1} a_k = infty$$
If I know what $b_k$ looks like, then I guess it is not hard to find out how to choose $a_k$. However we don't know what $b_k$ looks like and neither in which rate it decays. I really don't know what to do except that I have tried something like choosing for $n>m$
begin{align}
sum_{k=m}^n a_k = frac 1 {sqrt {b_mb_n^varepsilon}}
end{align}
for some $varepsilon>0$, so that the sequence is not Cauchy, that leads to the divergence of $sum_k a_k$ which is nice. At the same time I get
begin{align}
sum_{k=m}^n b_ka_k leq b_m sum_{k=m}^na_k =sqrt{frac{b_m}{b_n^varepsilon}}
end{align}
This tells us that, if there is $varepsilon$ such that
$$sqrt{frac{b_m}{b_n^varepsilon}}to 0 text{ as } n,mtoinfty$$
we are done. However, I don't think that will work, since $varepsilon$ is fixed...
Question. How can the problem be solved?
real-analysis sequences-and-series cauchy-sequences
real-analysis sequences-and-series cauchy-sequences
edited Jan 10 at 19:50
David Mitra
63.1k6100164
asked Jan 10 at 19:22
ShashiShashi
7,1731528
edited Jan 10 at 19:50
David Mitra
63.1k6100164
asked Jan 10 at 19:22
ShashiShashi
7,1731528
edited Jan 10 at 19:50
David Mitra
63.1k6100164
edited Jan 10 at 19:50
David Mitra
63.1k6100164
edited Jan 10 at 19:50
David Mitra
63.1k6100164
63.1k6100164
asked Jan 10 at 19:22
ShashiShashi
7,1731528
asked Jan 10 at 19:22
ShashiShashi
7,1731528
asked Jan 10 at 19:22
ShashiShashi
7,1731528
7,1731528
add a comment |
add a comment |
2 Answers
2
active
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votes
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Since $b_kto 0$, there is a strictly increasing subsequence $(k_n)_n$ such that
$$|b_{k_n}|<frac{1}{2^{n}}.$$
Now, for any $kin mathbb{N}$, let
$$a_k=begin{cases}1 &text{if $k=k_n$}\ 0 &text{otherwise}end{cases}.$$
answered Jan 10 at 19:30
Robert ZRobert Z
95.2k1063134
$endgroup$
$begingroup$
You didn't even use the fact that $b_n$ is decreasing.
$endgroup$
– N. S.
Jan 10 at 20:39
add a comment |
$begingroup$
Define $$a_k=begin{cases}1&,quad b_k<{1over 2^k},notexists u<k,b_{u}<{1over 2^{k}}\0&,quad text{elsewhere}end{cases}$$for example if $b_n={1over n}$ then we have $$a_n=begin{cases}1&,quad n=1,2,4,8,16,cdots \0&,quad text{elsewhere}end{cases}$$
answered Jan 10 at 19:41
Mostafa AyazMostafa Ayaz
15.3k3939
$endgroup$
add a comment |
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2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
Since $b_kto 0$, there is a strictly increasing subsequence $(k_n)_n$ such that
$$|b_{k_n}|<frac{1}{2^{n}}.$$
Now, for any $kin mathbb{N}$, let
$$a_k=begin{cases}1 &text{if $k=k_n$}\ 0 &text{otherwise}end{cases}.$$
answered Jan 10 at 19:30
Robert ZRobert Z
95.2k1063134
$endgroup$
$begingroup$
You didn't even use the fact that $b_n$ is decreasing.
$endgroup$
– N. S.
Jan 10 at 20:39
add a comment |
$begingroup$
Since $b_kto 0$, there is a strictly increasing subsequence $(k_n)_n$ such that
$$|b_{k_n}|<frac{1}{2^{n}}.$$
Now, for any $kin mathbb{N}$, let
$$a_k=begin{cases}1 &text{if $k=k_n$}\ 0 &text{otherwise}end{cases}.$$
answered Jan 10 at 19:30
Robert ZRobert Z
95.2k1063134
$endgroup$
$begingroup$
You didn't even use the fact that $b_n$ is decreasing.
$endgroup$
– N. S.
Jan 10 at 20:39
add a comment |
$begingroup$
Since $b_kto 0$, there is a strictly increasing subsequence $(k_n)_n$ such that
$$|b_{k_n}|<frac{1}{2^{n}}.$$
Now, for any $kin mathbb{N}$, let
$$a_k=begin{cases}1 &text{if $k=k_n$}\ 0 &text{otherwise}end{cases}.$$
answered Jan 10 at 19:30
Robert ZRobert Z
95.2k1063134
$endgroup$
Since $b_kto 0$, there is a strictly increasing subsequence $(k_n)_n$ such that
$$|b_{k_n}|<frac{1}{2^{n}}.$$
Now, for any $kin mathbb{N}$, let
$$a_k=begin{cases}1 &text{if $k=k_n$}\ 0 &text{otherwise}end{cases}.$$
answered Jan 10 at 19:30
Robert ZRobert Z
95.2k1063134
edited Jan 10 at 19:35
answered Jan 10 at 19:30
Robert ZRobert Z
95.2k1063134
answered Jan 10 at 19:30
Robert ZRobert Z
95.2k1063134
answered Jan 10 at 19:30
Robert ZRobert Z
95.2k1063134
95.2k1063134
$begingroup$
You didn't even use the fact that $b_n$ is decreasing.
$endgroup$
– N. S.
Jan 10 at 20:39
add a comment |
$begingroup$
You didn't even use the fact that $b_n$ is decreasing.
$endgroup$
– N. S.
Jan 10 at 20:39
$begingroup$
You didn't even use the fact that $b_n$ is decreasing.
$endgroup$
– N. S.
Jan 10 at 20:39
$begingroup$
You didn't even use the fact that $b_n$ is decreasing.
$endgroup$
– N. S.
Jan 10 at 20:39
add a comment |
$begingroup$
Define $$a_k=begin{cases}1&,quad b_k<{1over 2^k},notexists u<k,b_{u}<{1over 2^{k}}\0&,quad text{elsewhere}end{cases}$$for example if $b_n={1over n}$ then we have $$a_n=begin{cases}1&,quad n=1,2,4,8,16,cdots \0&,quad text{elsewhere}end{cases}$$
answered Jan 10 at 19:41
Mostafa AyazMostafa Ayaz
15.3k3939
$endgroup$
add a comment |
$begingroup$
Define $$a_k=begin{cases}1&,quad b_k<{1over 2^k},notexists u<k,b_{u}<{1over 2^{k}}\0&,quad text{elsewhere}end{cases}$$for example if $b_n={1over n}$ then we have $$a_n=begin{cases}1&,quad n=1,2,4,8,16,cdots \0&,quad text{elsewhere}end{cases}$$
answered Jan 10 at 19:41
Mostafa AyazMostafa Ayaz
15.3k3939
$endgroup$
add a comment |
$begingroup$
Define $$a_k=begin{cases}1&,quad b_k<{1over 2^k},notexists u<k,b_{u}<{1over 2^{k}}\0&,quad text{elsewhere}end{cases}$$for example if $b_n={1over n}$ then we have $$a_n=begin{cases}1&,quad n=1,2,4,8,16,cdots \0&,quad text{elsewhere}end{cases}$$
answered Jan 10 at 19:41
Mostafa AyazMostafa Ayaz
15.3k3939
$endgroup$
Define $$a_k=begin{cases}1&,quad b_k<{1over 2^k},notexists u<k,b_{u}<{1over 2^{k}}\0&,quad text{elsewhere}end{cases}$$for example if $b_n={1over n}$ then we have $$a_n=begin{cases}1&,quad n=1,2,4,8,16,cdots \0&,quad text{elsewhere}end{cases}$$
answered Jan 10 at 19:41
Mostafa AyazMostafa Ayaz
15.3k3939
edited Jan 10 at 19:48
answered Jan 10 at 19:41
Mostafa AyazMostafa Ayaz
15.3k3939
answered Jan 10 at 19:41
Mostafa AyazMostafa Ayaz
15.3k3939
answered Jan 10 at 19:41
Mostafa AyazMostafa Ayaz
15.3k3939
15.3k3939
add a comment |
add a comment |
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