On convergent series - in the spirit of Abel and Dini
Nonexistence of boundary between convergent and divergent series?
I'm hoping the following is true:
Suppose $a_i $ is a positive sequence and $sum_i a_i < infty.$ Then there exists a positive sequence $b_i$ s.t $sum_i b_i < infty$ and $sum_i frac{a_i}{b_i} < infty$.
sequences-and-series
asked Dec 1 at 14:43
Better2BLucky
393
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Nonexistence of boundary between convergent and divergent series?
I'm hoping the following is true:
Suppose $a_i $ is a positive sequence and $sum_i a_i < infty.$ Then there exists a positive sequence $b_i$ s.t $sum_i b_i < infty$ and $sum_i frac{a_i}{b_i} < infty$.
sequences-and-series
asked Dec 1 at 14:43
Better2BLucky
393
add a comment |
Nonexistence of boundary between convergent and divergent series?
I'm hoping the following is true:
Suppose $a_i $ is a positive sequence and $sum_i a_i < infty.$ Then there exists a positive sequence $b_i$ s.t $sum_i b_i < infty$ and $sum_i frac{a_i}{b_i} < infty$.
sequences-and-series
asked Dec 1 at 14:43
Better2BLucky
393
Nonexistence of boundary between convergent and divergent series?
I'm hoping the following is true:
Suppose $a_i $ is a positive sequence and $sum_i a_i < infty.$ Then there exists a positive sequence $b_i$ s.t $sum_i b_i < infty$ and $sum_i frac{a_i}{b_i} < infty$.
sequences-and-series
sequences-and-series
asked Dec 1 at 14:43
Better2BLucky
393
asked Dec 1 at 14:43
Better2BLucky
393
asked Dec 1 at 14:43
Better2BLucky
393
asked Dec 1 at 14:43
Better2BLucky
393
asked Dec 1 at 14:43
Better2BLucky
393
393
add a comment |
add a comment |
1 Answer
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This is not correct. Take $a_n=1/n^2$.
Let us show that $b_n$ with
required properties does not exist. Consider the set
$$E={ n: b_ngeq 1/n}.$$
As $sum b_n<infty$, we have $$sum_E1/n<infty.$$
Now on $Nbackslash E$ we have $b_n<1/n,$ so $1/b_n>n$ and as $sum a_n/b_n<infty$,
we conclude that $$sum_{Nbackslash E}1/n<infty.$$
adding the last two inequalities we obtain a contradiction.
answered Dec 1 at 15:03
Alexandre Eremenko
49k6136253
add a comment |
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1 Answer
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1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
This is not correct. Take $a_n=1/n^2$.
Let us show that $b_n$ with
required properties does not exist. Consider the set
$$E={ n: b_ngeq 1/n}.$$
As $sum b_n<infty$, we have $$sum_E1/n<infty.$$
Now on $Nbackslash E$ we have $b_n<1/n,$ so $1/b_n>n$ and as $sum a_n/b_n<infty$,
we conclude that $$sum_{Nbackslash E}1/n<infty.$$
adding the last two inequalities we obtain a contradiction.
answered Dec 1 at 15:03
Alexandre Eremenko
49k6136253
add a comment |
This is not correct. Take $a_n=1/n^2$.
Let us show that $b_n$ with
required properties does not exist. Consider the set
$$E={ n: b_ngeq 1/n}.$$
As $sum b_n<infty$, we have $$sum_E1/n<infty.$$
Now on $Nbackslash E$ we have $b_n<1/n,$ so $1/b_n>n$ and as $sum a_n/b_n<infty$,
we conclude that $$sum_{Nbackslash E}1/n<infty.$$
adding the last two inequalities we obtain a contradiction.
answered Dec 1 at 15:03
Alexandre Eremenko
49k6136253
add a comment |
This is not correct. Take $a_n=1/n^2$.
Let us show that $b_n$ with
required properties does not exist. Consider the set
$$E={ n: b_ngeq 1/n}.$$
As $sum b_n<infty$, we have $$sum_E1/n<infty.$$
Now on $Nbackslash E$ we have $b_n<1/n,$ so $1/b_n>n$ and as $sum a_n/b_n<infty$,
we conclude that $$sum_{Nbackslash E}1/n<infty.$$
adding the last two inequalities we obtain a contradiction.
answered Dec 1 at 15:03
Alexandre Eremenko
49k6136253
This is not correct. Take $a_n=1/n^2$.
Let us show that $b_n$ with
required properties does not exist. Consider the set
$$E={ n: b_ngeq 1/n}.$$
As $sum b_n<infty$, we have $$sum_E1/n<infty.$$
Now on $Nbackslash E$ we have $b_n<1/n,$ so $1/b_n>n$ and as $sum a_n/b_n<infty$,
we conclude that $$sum_{Nbackslash E}1/n<infty.$$
adding the last two inequalities we obtain a contradiction.
answered Dec 1 at 15:03
Alexandre Eremenko
49k6136253
answered Dec 1 at 15:03
Alexandre Eremenko
49k6136253
answered Dec 1 at 15:03
Alexandre Eremenko
49k6136253
answered Dec 1 at 15:03
Alexandre Eremenko
49k6136253
49k6136253
add a comment |
add a comment |
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