Is there integer functions $f(n), g(n)$, such that $lim_{n to infty} f(n)zeta(2) + g(n) = 0$












0












$begingroup$


Is there integer functions $f(n)$ and $g(n)$, such that $lim_{n to infty} f(n)zeta(2) + g(n) = 0$ where $f(n)zeta(2) + g(n) neq 0$ for all positive integer $n$.



For example:



$(-1)^n!n e + (-1)^{n+1}n! neq 0$ for all positive integer $n$ and the limit at infinity is $0$. Here $f(n) = (-1)^n!n$ and $g(n) = (-1)^{n+1}n!$ are integers functions for all positive integer $n$. Instead of $e$ I want to use $zeta(2)$.






number-theory







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$endgroup$












  • $begingroup$
    Equivalently, you want $zeta(2) approx -g(n)/f(n)$ where $g$ and $f$ are integers. There are arbitrarily good rational approximations, and so the answer is "yes" to your question.
    $endgroup$
    – davidlowryduda♦
    Dec 16 '18 at 5:12






  • 1




    $begingroup$
    @davidlowryduda $zeta(2) +g(n)/f(n)=o(1/f(n))$ is much stronger than $zeta(2) +g(n)/f(n)=o(1)$
    $endgroup$
    – reuns
    Dec 16 '18 at 5:22
















0












$begingroup$


Is there integer functions $f(n)$ and $g(n)$, such that $lim_{n to infty} f(n)zeta(2) + g(n) = 0$ where $f(n)zeta(2) + g(n) neq 0$ for all positive integer $n$.



For example:



$(-1)^n!n e + (-1)^{n+1}n! neq 0$ for all positive integer $n$ and the limit at infinity is $0$. Here $f(n) = (-1)^n!n$ and $g(n) = (-1)^{n+1}n!$ are integers functions for all positive integer $n$. Instead of $e$ I want to use $zeta(2)$.






number-theory







share|cite|improve this question









$endgroup$












  • $begingroup$
    Equivalently, you want $zeta(2) approx -g(n)/f(n)$ where $g$ and $f$ are integers. There are arbitrarily good rational approximations, and so the answer is "yes" to your question.
    $endgroup$
    – davidlowryduda♦
    Dec 16 '18 at 5:12






  • 1




    $begingroup$
    @davidlowryduda $zeta(2) +g(n)/f(n)=o(1/f(n))$ is much stronger than $zeta(2) +g(n)/f(n)=o(1)$
    $endgroup$
    – reuns
    Dec 16 '18 at 5:22














0












0








0





$begingroup$


Is there integer functions $f(n)$ and $g(n)$, such that $lim_{n to infty} f(n)zeta(2) + g(n) = 0$ where $f(n)zeta(2) + g(n) neq 0$ for all positive integer $n$.



For example:



$(-1)^n!n e + (-1)^{n+1}n! neq 0$ for all positive integer $n$ and the limit at infinity is $0$. Here $f(n) = (-1)^n!n$ and $g(n) = (-1)^{n+1}n!$ are integers functions for all positive integer $n$. Instead of $e$ I want to use $zeta(2)$.






number-theory







share|cite|improve this question









$endgroup$




Is there integer functions $f(n)$ and $g(n)$, such that $lim_{n to infty} f(n)zeta(2) + g(n) = 0$ where $f(n)zeta(2) + g(n) neq 0$ for all positive integer $n$.



For example:



$(-1)^n!n e + (-1)^{n+1}n! neq 0$ for all positive integer $n$ and the limit at infinity is $0$. Here $f(n) = (-1)^n!n$ and $g(n) = (-1)^{n+1}n!$ are integers functions for all positive integer $n$. Instead of $e$ I want to use $zeta(2)$.






number-theory




number-theory






share|cite|improve this question













share|cite|improve this question











share|cite|improve this question




share|cite|improve this question










asked Dec 16 '18 at 5:06









PintecoPinteco

731313




731313












  • $begingroup$
    Equivalently, you want $zeta(2) approx -g(n)/f(n)$ where $g$ and $f$ are integers. There are arbitrarily good rational approximations, and so the answer is "yes" to your question.
    $endgroup$
    – davidlowryduda♦
    Dec 16 '18 at 5:12






  • 1




    $begingroup$
    @davidlowryduda $zeta(2) +g(n)/f(n)=o(1/f(n))$ is much stronger than $zeta(2) +g(n)/f(n)=o(1)$
    $endgroup$
    – reuns
    Dec 16 '18 at 5:22


















  • $begingroup$
    Equivalently, you want $zeta(2) approx -g(n)/f(n)$ where $g$ and $f$ are integers. There are arbitrarily good rational approximations, and so the answer is "yes" to your question.
    $endgroup$
    – davidlowryduda♦
    Dec 16 '18 at 5:12






  • 1




    $begingroup$
    @davidlowryduda $zeta(2) +g(n)/f(n)=o(1/f(n))$ is much stronger than $zeta(2) +g(n)/f(n)=o(1)$
    $endgroup$
    – reuns
    Dec 16 '18 at 5:22
















$begingroup$
Equivalently, you want $zeta(2) approx -g(n)/f(n)$ where $g$ and $f$ are integers. There are arbitrarily good rational approximations, and so the answer is "yes" to your question.
$endgroup$
– davidlowryduda♦
Dec 16 '18 at 5:12




$begingroup$
Equivalently, you want $zeta(2) approx -g(n)/f(n)$ where $g$ and $f$ are integers. There are arbitrarily good rational approximations, and so the answer is "yes" to your question.
$endgroup$
– davidlowryduda♦
Dec 16 '18 at 5:12




1




1




$begingroup$
@davidlowryduda $zeta(2) +g(n)/f(n)=o(1/f(n))$ is much stronger than $zeta(2) +g(n)/f(n)=o(1)$
$endgroup$
– reuns
Dec 16 '18 at 5:22




$begingroup$
@davidlowryduda $zeta(2) +g(n)/f(n)=o(1/f(n))$ is much stronger than $zeta(2) +g(n)/f(n)=o(1)$
$endgroup$
– reuns
Dec 16 '18 at 5:22










1 Answer
1






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oldest

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$begingroup$

For any real number $x$, there exist functions $f,g$ so that



$$lim_{ntoinfty} xf(n)+g(n)=0.$$



If $x$ is rational, then this is trivial. Otherwise, let $frac{p_n}{q_n}$ be the $n$th convergent to the continued fraction of $x$, and let $g(n)=-p_n$, $f(n)=q_n$.



For more about continued fraction convergents, see Wikipedia.






share|cite|improve this answer











$endgroup$













  • $begingroup$
    It's possible that these $f,g$ are integer valued functions for all integer $n$? If so, how to find they?
    $endgroup$
    – Pinteco
    Dec 16 '18 at 5:13










  • $begingroup$
    @Pinteco Yes, continued fraction convergents are always integers, as they're all about approximating real numbers by rational numbers. I've added a link about them in my answer.
    $endgroup$
    – Carl Schildkraut
    Dec 16 '18 at 5:18











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1 Answer
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1 Answer
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2












$begingroup$

For any real number $x$, there exist functions $f,g$ so that



$$lim_{ntoinfty} xf(n)+g(n)=0.$$



If $x$ is rational, then this is trivial. Otherwise, let $frac{p_n}{q_n}$ be the $n$th convergent to the continued fraction of $x$, and let $g(n)=-p_n$, $f(n)=q_n$.



For more about continued fraction convergents, see Wikipedia.






share|cite|improve this answer











$endgroup$













  • $begingroup$
    It's possible that these $f,g$ are integer valued functions for all integer $n$? If so, how to find they?
    $endgroup$
    – Pinteco
    Dec 16 '18 at 5:13










  • $begingroup$
    @Pinteco Yes, continued fraction convergents are always integers, as they're all about approximating real numbers by rational numbers. I've added a link about them in my answer.
    $endgroup$
    – Carl Schildkraut
    Dec 16 '18 at 5:18
















2












$begingroup$

For any real number $x$, there exist functions $f,g$ so that



$$lim_{ntoinfty} xf(n)+g(n)=0.$$



If $x$ is rational, then this is trivial. Otherwise, let $frac{p_n}{q_n}$ be the $n$th convergent to the continued fraction of $x$, and let $g(n)=-p_n$, $f(n)=q_n$.



For more about continued fraction convergents, see Wikipedia.






share|cite|improve this answer











$endgroup$













  • $begingroup$
    It's possible that these $f,g$ are integer valued functions for all integer $n$? If so, how to find they?
    $endgroup$
    – Pinteco
    Dec 16 '18 at 5:13










  • $begingroup$
    @Pinteco Yes, continued fraction convergents are always integers, as they're all about approximating real numbers by rational numbers. I've added a link about them in my answer.
    $endgroup$
    – Carl Schildkraut
    Dec 16 '18 at 5:18














2












2








2





$begingroup$

For any real number $x$, there exist functions $f,g$ so that



$$lim_{ntoinfty} xf(n)+g(n)=0.$$



If $x$ is rational, then this is trivial. Otherwise, let $frac{p_n}{q_n}$ be the $n$th convergent to the continued fraction of $x$, and let $g(n)=-p_n$, $f(n)=q_n$.



For more about continued fraction convergents, see Wikipedia.






share|cite|improve this answer











$endgroup$



For any real number $x$, there exist functions $f,g$ so that



$$lim_{ntoinfty} xf(n)+g(n)=0.$$



If $x$ is rational, then this is trivial. Otherwise, let $frac{p_n}{q_n}$ be the $n$th convergent to the continued fraction of $x$, and let $g(n)=-p_n$, $f(n)=q_n$.



For more about continued fraction convergents, see Wikipedia.







share|cite|improve this answer














share|cite|improve this answer



share|cite|improve this answer








edited Dec 16 '18 at 5:18

























answered Dec 16 '18 at 5:10









Carl SchildkrautCarl Schildkraut

11.7k11443




11.7k11443












  • $begingroup$
    It's possible that these $f,g$ are integer valued functions for all integer $n$? If so, how to find they?
    $endgroup$
    – Pinteco
    Dec 16 '18 at 5:13










  • $begingroup$
    @Pinteco Yes, continued fraction convergents are always integers, as they're all about approximating real numbers by rational numbers. I've added a link about them in my answer.
    $endgroup$
    – Carl Schildkraut
    Dec 16 '18 at 5:18


















  • $begingroup$
    It's possible that these $f,g$ are integer valued functions for all integer $n$? If so, how to find they?
    $endgroup$
    – Pinteco
    Dec 16 '18 at 5:13










  • $begingroup$
    @Pinteco Yes, continued fraction convergents are always integers, as they're all about approximating real numbers by rational numbers. I've added a link about them in my answer.
    $endgroup$
    – Carl Schildkraut
    Dec 16 '18 at 5:18
















$begingroup$
It's possible that these $f,g$ are integer valued functions for all integer $n$? If so, how to find they?
$endgroup$
– Pinteco
Dec 16 '18 at 5:13




$begingroup$
It's possible that these $f,g$ are integer valued functions for all integer $n$? If so, how to find they?
$endgroup$
– Pinteco
Dec 16 '18 at 5:13












$begingroup$
@Pinteco Yes, continued fraction convergents are always integers, as they're all about approximating real numbers by rational numbers. I've added a link about them in my answer.
$endgroup$
– Carl Schildkraut
Dec 16 '18 at 5:18




$begingroup$
@Pinteco Yes, continued fraction convergents are always integers, as they're all about approximating real numbers by rational numbers. I've added a link about them in my answer.
$endgroup$
– Carl Schildkraut
Dec 16 '18 at 5:18


















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