Mixing
Checking whether given series is nonnegative.
$begingroup$
I have a question: is this true that
$$sum_{n=1}^{infty}a(n)left(frac{1}{n^{a}}-frac{1}{n^{b}}right)^{2}ge 0$$
For every $a,b in (0,0.5)$?
Where $a(n)=sum_{d|n}(-1)^{d+frac{n}{d}}cosleft(ylnleft(frac{n}{d^{2}}right)right)$
I tried to use Abel's summation formula or summation by parts, but i have no idea how to solve this problem.
Also i need some hints how to generally solve that kind of problems (i mean proving that given infinite series is nonnegative).
I will be grateful if someone help me.
Regards.
real-analysis sequences-and-series
edited Jan 16 at 21:15
user3482749
4,266919
asked Jan 16 at 20:55
mkultramkultra
708
$endgroup$
add a comment |
$begingroup$
I have a question: is this true that
$$sum_{n=1}^{infty}a(n)left(frac{1}{n^{a}}-frac{1}{n^{b}}right)^{2}ge 0$$
For every $a,b in (0,0.5)$?
Where $a(n)=sum_{d|n}(-1)^{d+frac{n}{d}}cosleft(ylnleft(frac{n}{d^{2}}right)right)$
I tried to use Abel's summation formula or summation by parts, but i have no idea how to solve this problem.
Also i need some hints how to generally solve that kind of problems (i mean proving that given infinite series is nonnegative).
I will be grateful if someone help me.
Regards.
real-analysis sequences-and-series
edited Jan 16 at 21:15
user3482749
4,266919
asked Jan 16 at 20:55
mkultramkultra
708
$endgroup$
$begingroup$
What is $y$, here?
$endgroup$
– user3482749
Jan 16 at 21:17
$begingroup$
It is a positive real number.
$endgroup$
– mkultra
Jan 16 at 21:25
$begingroup$
In that case, it strikes me as exceedingly unlikely that this is true in general. For a method of proving it: take the maximally-bad $y$ to make the first few terms as negative as possible, then bound the size of the remaining terms in terms of that. Working on an explicit bounding now.
$endgroup$
– user3482749
Jan 16 at 21:29
$begingroup$
There is one more question. Is it true that: $$sum_{n=1}^{infty}a(n)frac{(ln n)^{2}}{n^x}ge 0$$ For all $x>0$ ?
$endgroup$
– mkultra
Jan 19 at 11:14
add a comment |
$begingroup$
I have a question: is this true that
$$sum_{n=1}^{infty}a(n)left(frac{1}{n^{a}}-frac{1}{n^{b}}right)^{2}ge 0$$
For every $a,b in (0,0.5)$?
Where $a(n)=sum_{d|n}(-1)^{d+frac{n}{d}}cosleft(ylnleft(frac{n}{d^{2}}right)right)$
I tried to use Abel's summation formula or summation by parts, but i have no idea how to solve this problem.
Also i need some hints how to generally solve that kind of problems (i mean proving that given infinite series is nonnegative).
I will be grateful if someone help me.
Regards.
real-analysis sequences-and-series
edited Jan 16 at 21:15
user3482749
4,266919
asked Jan 16 at 20:55
mkultramkultra
708
$endgroup$
I have a question: is this true that
$$sum_{n=1}^{infty}a(n)left(frac{1}{n^{a}}-frac{1}{n^{b}}right)^{2}ge 0$$
For every $a,b in (0,0.5)$?
Where $a(n)=sum_{d|n}(-1)^{d+frac{n}{d}}cosleft(ylnleft(frac{n}{d^{2}}right)right)$
I tried to use Abel's summation formula or summation by parts, but i have no idea how to solve this problem.
Also i need some hints how to generally solve that kind of problems (i mean proving that given infinite series is nonnegative).
I will be grateful if someone help me.
Regards.
real-analysis sequences-and-series
real-analysis sequences-and-series
edited Jan 16 at 21:15
user3482749
4,266919
asked Jan 16 at 20:55
mkultramkultra
708
edited Jan 16 at 21:15
user3482749
4,266919
asked Jan 16 at 20:55
mkultramkultra
708
edited Jan 16 at 21:15
user3482749
4,266919
edited Jan 16 at 21:15
user3482749
4,266919
edited Jan 16 at 21:15
user3482749
4,266919
4,266919
asked Jan 16 at 20:55
mkultramkultra
708
asked Jan 16 at 20:55
mkultramkultra
708
asked Jan 16 at 20:55
mkultramkultra
708
708
$begingroup$
What is $y$, here?
$endgroup$
– user3482749
Jan 16 at 21:17
$begingroup$
It is a positive real number.
$endgroup$
– mkultra
Jan 16 at 21:25
$begingroup$
In that case, it strikes me as exceedingly unlikely that this is true in general. For a method of proving it: take the maximally-bad $y$ to make the first few terms as negative as possible, then bound the size of the remaining terms in terms of that. Working on an explicit bounding now.
$endgroup$
– user3482749
Jan 16 at 21:29
$begingroup$
There is one more question. Is it true that: $$sum_{n=1}^{infty}a(n)frac{(ln n)^{2}}{n^x}ge 0$$ For all $x>0$ ?
$endgroup$
– mkultra
Jan 19 at 11:14
add a comment |
$begingroup$
What is $y$, here?
$endgroup$
– user3482749
Jan 16 at 21:17
$begingroup$
It is a positive real number.
$endgroup$
– mkultra
Jan 16 at 21:25
$begingroup$
In that case, it strikes me as exceedingly unlikely that this is true in general. For a method of proving it: take the maximally-bad $y$ to make the first few terms as negative as possible, then bound the size of the remaining terms in terms of that. Working on an explicit bounding now.
$endgroup$
– user3482749
Jan 16 at 21:29
$begingroup$
There is one more question. Is it true that: $$sum_{n=1}^{infty}a(n)frac{(ln n)^{2}}{n^x}ge 0$$ For all $x>0$ ?
$endgroup$
– mkultra
Jan 19 at 11:14
$begingroup$
What is $y$, here?
$endgroup$
– user3482749
Jan 16 at 21:17
$begingroup$
What is $y$, here?
$endgroup$
– user3482749
Jan 16 at 21:17
$begingroup$
It is a positive real number.
$endgroup$
– mkultra
Jan 16 at 21:25
$begingroup$
It is a positive real number.
$endgroup$
– mkultra
Jan 16 at 21:25
$begingroup$
In that case, it strikes me as exceedingly unlikely that this is true in general. For a method of proving it: take the maximally-bad $y$ to make the first few terms as negative as possible, then bound the size of the remaining terms in terms of that. Working on an explicit bounding now.
$endgroup$
– user3482749
Jan 16 at 21:29
$begingroup$
In that case, it strikes me as exceedingly unlikely that this is true in general. For a method of proving it: take the maximally-bad $y$ to make the first few terms as negative as possible, then bound the size of the remaining terms in terms of that. Working on an explicit bounding now.
$endgroup$
– user3482749
Jan 16 at 21:29
$begingroup$
There is one more question. Is it true that: $$sum_{n=1}^{infty}a(n)frac{(ln n)^{2}}{n^x}ge 0$$ For all $x>0$ ?
$endgroup$
– mkultra
Jan 19 at 11:14
$begingroup$
There is one more question. Is it true that: $$sum_{n=1}^{infty}a(n)frac{(ln n)^{2}}{n^x}ge 0$$ For all $x>0$ ?
$endgroup$
– mkultra
Jan 19 at 11:14
add a comment |
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$begingroup$
What is $y$, here?
$endgroup$
– user3482749
Jan 16 at 21:17
$begingroup$
It is a positive real number.
$endgroup$
– mkultra
Jan 16 at 21:25
$begingroup$
In that case, it strikes me as exceedingly unlikely that this is true in general. For a method of proving it: take the maximally-bad $y$ to make the first few terms as negative as possible, then bound the size of the remaining terms in terms of that. Working on an explicit bounding now.
$endgroup$
– user3482749
Jan 16 at 21:29
$begingroup$
There is one more question. Is it true that: $$sum_{n=1}^{infty}a(n)frac{(ln n)^{2}}{n^x}ge 0$$ For all $x>0$ ?
$endgroup$
– mkultra
Jan 19 at 11:14