Convergence of a special sequence.
$begingroup$
There is a sequence:
$$a_{n}=sum_{d|n}(-1)^{d+frac{n}{d}}cos(ln(frac{n}{d^{2}}))$$
Is this true that $a_{n}rightarrow0$ ?
It seems that this sequence rather diverges.
I tried to prove its discrepancy by looking for special divergent subsequences, but i can't find.
I hope that someone will help me.
Regards.
sequences-and-series number-theory convergence
asked Jan 13 at 19:38
mkultramkultra
568
$endgroup$
add a comment |
$begingroup$
There is a sequence:
$$a_{n}=sum_{d|n}(-1)^{d+frac{n}{d}}cos(ln(frac{n}{d^{2}}))$$
Is this true that $a_{n}rightarrow0$ ?
It seems that this sequence rather diverges.
I tried to prove its discrepancy by looking for special divergent subsequences, but i can't find.
I hope that someone will help me.
Regards.
sequences-and-series number-theory convergence
asked Jan 13 at 19:38
mkultramkultra
568
$endgroup$
add a comment |
$begingroup$
There is a sequence:
$$a_{n}=sum_{d|n}(-1)^{d+frac{n}{d}}cos(ln(frac{n}{d^{2}}))$$
Is this true that $a_{n}rightarrow0$ ?
It seems that this sequence rather diverges.
I tried to prove its discrepancy by looking for special divergent subsequences, but i can't find.
I hope that someone will help me.
Regards.
sequences-and-series number-theory convergence
asked Jan 13 at 19:38
mkultramkultra
568
$endgroup$
There is a sequence:
$$a_{n}=sum_{d|n}(-1)^{d+frac{n}{d}}cos(ln(frac{n}{d^{2}}))$$
Is this true that $a_{n}rightarrow0$ ?
It seems that this sequence rather diverges.
I tried to prove its discrepancy by looking for special divergent subsequences, but i can't find.
I hope that someone will help me.
Regards.
sequences-and-series number-theory convergence
sequences-and-series number-theory convergence
asked Jan 13 at 19:38
mkultramkultra
568
asked Jan 13 at 19:38
mkultramkultra
568
asked Jan 13 at 19:38
mkultramkultra
568
asked Jan 13 at 19:38
mkultramkultra
568
asked Jan 13 at 19:38
mkultramkultra
568
568
add a comment |
add a comment |
1 Answer
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If $p$ is an odd prime, then $$a_p=(-1)^{p+1}(cos(ln(p))+cos(ln(1/p))=2cos(ln(p)).$$ By Bertrand's postulate, the logarithms of consecutive primes are within $ln(2)<pi/4$ of each other. It follows that every interval of the form $[Npi/4,(N+1)pi/4]$ for $N$ a positive integer contains a number of the form $ln(p)$ for some prime $p$. In particular, this means that $cos(ln(p))$ takes both values greater than $1/sqrt{2}$ and values less than $-1/sqrt{2}$ infinitely often. It follows that the sequence $(a_p)$ cannot converge as $p$ ranges over the primes.
answered Jan 13 at 19:50
Eric WofseyEric Wofsey
184k13212338
$endgroup$
$begingroup$
But what if i modify a little bit and put $a_{p}=2cos(yln(p))$, for $y>0$ ?
$endgroup$
– mkultra
Jan 13 at 22:40
$begingroup$
Yes: the ratio between consecutive primes converges to $1$ and so the difference of their logarithms converges to $0$ and you can make a similar argument for any $y>0$. This is a consequence of the prime number theorem (see en.wikipedia.org/wiki/Bertrand%27s_postulate#Better_results for instance).
$endgroup$
– Eric Wofsey
Jan 13 at 22:49
add a comment |
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1 Answer
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$begingroup$
If $p$ is an odd prime, then $$a_p=(-1)^{p+1}(cos(ln(p))+cos(ln(1/p))=2cos(ln(p)).$$ By Bertrand's postulate, the logarithms of consecutive primes are within $ln(2)<pi/4$ of each other. It follows that every interval of the form $[Npi/4,(N+1)pi/4]$ for $N$ a positive integer contains a number of the form $ln(p)$ for some prime $p$. In particular, this means that $cos(ln(p))$ takes both values greater than $1/sqrt{2}$ and values less than $-1/sqrt{2}$ infinitely often. It follows that the sequence $(a_p)$ cannot converge as $p$ ranges over the primes.
answered Jan 13 at 19:50
Eric WofseyEric Wofsey
184k13212338
$endgroup$
$begingroup$
But what if i modify a little bit and put $a_{p}=2cos(yln(p))$, for $y>0$ ?
$endgroup$
– mkultra
Jan 13 at 22:40
$begingroup$
Yes: the ratio between consecutive primes converges to $1$ and so the difference of their logarithms converges to $0$ and you can make a similar argument for any $y>0$. This is a consequence of the prime number theorem (see en.wikipedia.org/wiki/Bertrand%27s_postulate#Better_results for instance).
$endgroup$
– Eric Wofsey
Jan 13 at 22:49
add a comment |
$begingroup$
If $p$ is an odd prime, then $$a_p=(-1)^{p+1}(cos(ln(p))+cos(ln(1/p))=2cos(ln(p)).$$ By Bertrand's postulate, the logarithms of consecutive primes are within $ln(2)<pi/4$ of each other. It follows that every interval of the form $[Npi/4,(N+1)pi/4]$ for $N$ a positive integer contains a number of the form $ln(p)$ for some prime $p$. In particular, this means that $cos(ln(p))$ takes both values greater than $1/sqrt{2}$ and values less than $-1/sqrt{2}$ infinitely often. It follows that the sequence $(a_p)$ cannot converge as $p$ ranges over the primes.
answered Jan 13 at 19:50
Eric WofseyEric Wofsey
184k13212338
$endgroup$
$begingroup$
But what if i modify a little bit and put $a_{p}=2cos(yln(p))$, for $y>0$ ?
$endgroup$
– mkultra
Jan 13 at 22:40
$begingroup$
Yes: the ratio between consecutive primes converges to $1$ and so the difference of their logarithms converges to $0$ and you can make a similar argument for any $y>0$. This is a consequence of the prime number theorem (see en.wikipedia.org/wiki/Bertrand%27s_postulate#Better_results for instance).
$endgroup$
– Eric Wofsey
Jan 13 at 22:49
add a comment |
$begingroup$
If $p$ is an odd prime, then $$a_p=(-1)^{p+1}(cos(ln(p))+cos(ln(1/p))=2cos(ln(p)).$$ By Bertrand's postulate, the logarithms of consecutive primes are within $ln(2)<pi/4$ of each other. It follows that every interval of the form $[Npi/4,(N+1)pi/4]$ for $N$ a positive integer contains a number of the form $ln(p)$ for some prime $p$. In particular, this means that $cos(ln(p))$ takes both values greater than $1/sqrt{2}$ and values less than $-1/sqrt{2}$ infinitely often. It follows that the sequence $(a_p)$ cannot converge as $p$ ranges over the primes.
answered Jan 13 at 19:50
Eric WofseyEric Wofsey
184k13212338
$endgroup$
If $p$ is an odd prime, then $$a_p=(-1)^{p+1}(cos(ln(p))+cos(ln(1/p))=2cos(ln(p)).$$ By Bertrand's postulate, the logarithms of consecutive primes are within $ln(2)<pi/4$ of each other. It follows that every interval of the form $[Npi/4,(N+1)pi/4]$ for $N$ a positive integer contains a number of the form $ln(p)$ for some prime $p$. In particular, this means that $cos(ln(p))$ takes both values greater than $1/sqrt{2}$ and values less than $-1/sqrt{2}$ infinitely often. It follows that the sequence $(a_p)$ cannot converge as $p$ ranges over the primes.
answered Jan 13 at 19:50
Eric WofseyEric Wofsey
184k13212338
answered Jan 13 at 19:50
Eric WofseyEric Wofsey
184k13212338
answered Jan 13 at 19:50
Eric WofseyEric Wofsey
184k13212338
answered Jan 13 at 19:50
Eric WofseyEric Wofsey
184k13212338
184k13212338
$begingroup$
But what if i modify a little bit and put $a_{p}=2cos(yln(p))$, for $y>0$ ?
$endgroup$
– mkultra
Jan 13 at 22:40
$begingroup$
Yes: the ratio between consecutive primes converges to $1$ and so the difference of their logarithms converges to $0$ and you can make a similar argument for any $y>0$. This is a consequence of the prime number theorem (see en.wikipedia.org/wiki/Bertrand%27s_postulate#Better_results for instance).
$endgroup$
– Eric Wofsey
Jan 13 at 22:49
add a comment |
$begingroup$
But what if i modify a little bit and put $a_{p}=2cos(yln(p))$, for $y>0$ ?
$endgroup$
– mkultra
Jan 13 at 22:40
$begingroup$
Yes: the ratio between consecutive primes converges to $1$ and so the difference of their logarithms converges to $0$ and you can make a similar argument for any $y>0$. This is a consequence of the prime number theorem (see en.wikipedia.org/wiki/Bertrand%27s_postulate#Better_results for instance).
$endgroup$
– Eric Wofsey
Jan 13 at 22:49
$begingroup$
But what if i modify a little bit and put $a_{p}=2cos(yln(p))$, for $y>0$ ?
$endgroup$
– mkultra
Jan 13 at 22:40
$begingroup$
But what if i modify a little bit and put $a_{p}=2cos(yln(p))$, for $y>0$ ?
$endgroup$
– mkultra
Jan 13 at 22:40
$begingroup$
Yes: the ratio between consecutive primes converges to $1$ and so the difference of their logarithms converges to $0$ and you can make a similar argument for any $y>0$. This is a consequence of the prime number theorem (see en.wikipedia.org/wiki/Bertrand%27s_postulate#Better_results for instance).
$endgroup$
– Eric Wofsey
Jan 13 at 22:49
$begingroup$
Yes: the ratio between consecutive primes converges to $1$ and so the difference of their logarithms converges to $0$ and you can make a similar argument for any $y>0$. This is a consequence of the prime number theorem (see en.wikipedia.org/wiki/Bertrand%27s_postulate#Better_results for instance).
$endgroup$
– Eric Wofsey
Jan 13 at 22:49
add a comment |
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