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Is it known the value of $sumlimits_{n=1}^{infty}frac{log(n)}{ n^2}$?



Announcing the arrival of Valued Associate #679: Cesar Manara
Planned maintenance scheduled April 17/18, 2019 at 00:00UTC (8:00pm US/Eastern)Does the series $sumlimits_{n=1}^{infty}frac{sin(n-sqrt{n^2+n})}{n}$ converge?Finding the convergence interval of $sumlimits_{n=0}^{infty} frac{n!x^n}{n^n}$.Strategy for finding Maclaurin seriesFind the limit of $sum frac{1}{log^n(n)}$Radius and Interval of Convergence of $sumlimits_{n=1}^{infty}dfrac{x^n}{2n-1}$Write $sumlimits_{n=0}^infty e^{-xn^3}$ in the form $sumlimits_{n=-infty}^infty a_nx^n$Determine whether $sumlimits_{n=1}^{infty} (-1)^{n-1}(frac{n}{n^2+1})$ is absolutely convergent, conditionally convergent, or divergent.Computing:$sum_{n=0}^inftyfrac{3^n}{n!(n+3)}$Does $sumlimits_{n=1}^inftyfrac{1}{n^2}(a_1+cdots+ a_n)$ converges when $a_n$ converges?Finding the sum $sum_{n=0}^inftyfrac{(x+1)^{n+2}}{(n+2)!}$. I cannot use my usual methods that I am use to.












1












$begingroup$


In my information theory course, we have been asked to find the entropy of a particular distribution in $mathbb{N}$. In order to do so, I have come to the following integral
$$sumlimits_{n=1}^{infty}frac{lfloor log_2n rfloor }{2^{2lfloor log_2n rfloor}}$$
I would be content approximating it by $sumlimits_{n=1}^{infty}dfrac{log_2(n)}{ n^2}$, but I don't know how to compute it neither (or if this is even possible).



I know that the series $sumlimits_{n=1}^{infty}dfrac{log(n)}{ n^2}$ (with the natural logarithm for example), converges (Bertrand series), but I would like to know if its value is known.



I apologize if this has already been answered, or if it is easy to find somewhere else.






calculus summation information-theory







share|cite|improve this question











$endgroup$








  • 2




    $begingroup$
    It's $-zeta'(2)$, if that satisfies you.
    $endgroup$
    – Lord Shark the Unknown
    Mar 24 at 11:17










  • $begingroup$
    ... and $zeta'(2)$ you will get with the first derivation of $zeta(1-s)=frac{2}{(2pi)^s}cosfrac{pi s}{2}Gamma(s)zeta(s)$ at $s=2$ and $zeta'(-1)=frac{1}{2}-ln A$ where $A$ is called the Glaisher-Kinkelin constant, e.g. here. But it's better you look at the answer of GEdgar.
    $endgroup$
    – user90369
    Mar 24 at 11:49
















1












$begingroup$


In my information theory course, we have been asked to find the entropy of a particular distribution in $mathbb{N}$. In order to do so, I have come to the following integral
$$sumlimits_{n=1}^{infty}frac{lfloor log_2n rfloor }{2^{2lfloor log_2n rfloor}}$$
I would be content approximating it by $sumlimits_{n=1}^{infty}dfrac{log_2(n)}{ n^2}$, but I don't know how to compute it neither (or if this is even possible).



I know that the series $sumlimits_{n=1}^{infty}dfrac{log(n)}{ n^2}$ (with the natural logarithm for example), converges (Bertrand series), but I would like to know if its value is known.



I apologize if this has already been answered, or if it is easy to find somewhere else.






calculus summation information-theory







share|cite|improve this question











$endgroup$








  • 2




    $begingroup$
    It's $-zeta'(2)$, if that satisfies you.
    $endgroup$
    – Lord Shark the Unknown
    Mar 24 at 11:17










  • $begingroup$
    ... and $zeta'(2)$ you will get with the first derivation of $zeta(1-s)=frac{2}{(2pi)^s}cosfrac{pi s}{2}Gamma(s)zeta(s)$ at $s=2$ and $zeta'(-1)=frac{1}{2}-ln A$ where $A$ is called the Glaisher-Kinkelin constant, e.g. here. But it's better you look at the answer of GEdgar.
    $endgroup$
    – user90369
    Mar 24 at 11:49














1












1








1





$begingroup$


In my information theory course, we have been asked to find the entropy of a particular distribution in $mathbb{N}$. In order to do so, I have come to the following integral
$$sumlimits_{n=1}^{infty}frac{lfloor log_2n rfloor }{2^{2lfloor log_2n rfloor}}$$
I would be content approximating it by $sumlimits_{n=1}^{infty}dfrac{log_2(n)}{ n^2}$, but I don't know how to compute it neither (or if this is even possible).



I know that the series $sumlimits_{n=1}^{infty}dfrac{log(n)}{ n^2}$ (with the natural logarithm for example), converges (Bertrand series), but I would like to know if its value is known.



I apologize if this has already been answered, or if it is easy to find somewhere else.






calculus summation information-theory







share|cite|improve this question











$endgroup$




In my information theory course, we have been asked to find the entropy of a particular distribution in $mathbb{N}$. In order to do so, I have come to the following integral
$$sumlimits_{n=1}^{infty}frac{lfloor log_2n rfloor }{2^{2lfloor log_2n rfloor}}$$
I would be content approximating it by $sumlimits_{n=1}^{infty}dfrac{log_2(n)}{ n^2}$, but I don't know how to compute it neither (or if this is even possible).



I know that the series $sumlimits_{n=1}^{infty}dfrac{log(n)}{ n^2}$ (with the natural logarithm for example), converges (Bertrand series), but I would like to know if its value is known.



I apologize if this has already been answered, or if it is easy to find somewhere else.






calculus summation information-theory




calculus summation information-theory






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Mar 24 at 11:37









StubbornAtom

6,45431440




6,45431440










asked Mar 24 at 11:14









AcasAcas

6711




6711








  • 2




    $begingroup$
    It's $-zeta'(2)$, if that satisfies you.
    $endgroup$
    – Lord Shark the Unknown
    Mar 24 at 11:17










  • $begingroup$
    ... and $zeta'(2)$ you will get with the first derivation of $zeta(1-s)=frac{2}{(2pi)^s}cosfrac{pi s}{2}Gamma(s)zeta(s)$ at $s=2$ and $zeta'(-1)=frac{1}{2}-ln A$ where $A$ is called the Glaisher-Kinkelin constant, e.g. here. But it's better you look at the answer of GEdgar.
    $endgroup$
    – user90369
    Mar 24 at 11:49














  • 2




    $begingroup$
    It's $-zeta'(2)$, if that satisfies you.
    $endgroup$
    – Lord Shark the Unknown
    Mar 24 at 11:17










  • $begingroup$
    ... and $zeta'(2)$ you will get with the first derivation of $zeta(1-s)=frac{2}{(2pi)^s}cosfrac{pi s}{2}Gamma(s)zeta(s)$ at $s=2$ and $zeta'(-1)=frac{1}{2}-ln A$ where $A$ is called the Glaisher-Kinkelin constant, e.g. here. But it's better you look at the answer of GEdgar.
    $endgroup$
    – user90369
    Mar 24 at 11:49








2




2




$begingroup$
It's $-zeta'(2)$, if that satisfies you.
$endgroup$
– Lord Shark the Unknown
Mar 24 at 11:17




$begingroup$
It's $-zeta'(2)$, if that satisfies you.
$endgroup$
– Lord Shark the Unknown
Mar 24 at 11:17












$begingroup$
... and $zeta'(2)$ you will get with the first derivation of $zeta(1-s)=frac{2}{(2pi)^s}cosfrac{pi s}{2}Gamma(s)zeta(s)$ at $s=2$ and $zeta'(-1)=frac{1}{2}-ln A$ where $A$ is called the Glaisher-Kinkelin constant, e.g. here. But it's better you look at the answer of GEdgar.
$endgroup$
– user90369
Mar 24 at 11:49




$begingroup$
... and $zeta'(2)$ you will get with the first derivation of $zeta(1-s)=frac{2}{(2pi)^s}cosfrac{pi s}{2}Gamma(s)zeta(s)$ at $s=2$ and $zeta'(-1)=frac{1}{2}-ln A$ where $A$ is called the Glaisher-Kinkelin constant, e.g. here. But it's better you look at the answer of GEdgar.
$endgroup$
– user90369
Mar 24 at 11:49










1 Answer
1






active

oldest

votes


















9












$begingroup$

With the integer part still in there
$$
sumlimits_{n=1}^{infty}dfrac{lfloor log_2n rfloor }{2^{2lfloor log_2n rfloor}}
$$
proceed like this.



For $n=1$ we have $lfloor log_2n rfloor = 0$.



For $2 le n < 4$ we have $lfloor log_2n rfloor = 1.qquad$ (two terms)



For $4 le n < 8$ we have $lfloor log_2n rfloor = 2.qquad$ (four terms)



And so on,



For $2^k le n < 2^{k+1}$ we have $lfloor log_2n rfloor = k.qquad$ ($2^k$ terms)



So
$$
sumlimits_{n=1}^{infty}dfrac{lfloor log_2n rfloor }{2^{2lfloor log_2n rfloor}} =
sum_{k=0}^infty 2^k;frac{k}{2^{2k}} = 2
$$






share|cite|improve this answer









$endgroup$









  • 1




    $begingroup$
    This has been incredibly helpful, much simpler than expected. Thank you!
    $endgroup$
    – Acas
    Mar 24 at 12:09












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1 Answer
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active

oldest

votes








1 Answer
1






active

oldest

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active

oldest

votes






active

oldest

votes









9












$begingroup$

With the integer part still in there
$$
sumlimits_{n=1}^{infty}dfrac{lfloor log_2n rfloor }{2^{2lfloor log_2n rfloor}}
$$
proceed like this.



For $n=1$ we have $lfloor log_2n rfloor = 0$.



For $2 le n < 4$ we have $lfloor log_2n rfloor = 1.qquad$ (two terms)



For $4 le n < 8$ we have $lfloor log_2n rfloor = 2.qquad$ (four terms)



And so on,



For $2^k le n < 2^{k+1}$ we have $lfloor log_2n rfloor = k.qquad$ ($2^k$ terms)



So
$$
sumlimits_{n=1}^{infty}dfrac{lfloor log_2n rfloor }{2^{2lfloor log_2n rfloor}} =
sum_{k=0}^infty 2^k;frac{k}{2^{2k}} = 2
$$






share|cite|improve this answer









$endgroup$









  • 1




    $begingroup$
    This has been incredibly helpful, much simpler than expected. Thank you!
    $endgroup$
    – Acas
    Mar 24 at 12:09
















9












$begingroup$

With the integer part still in there
$$
sumlimits_{n=1}^{infty}dfrac{lfloor log_2n rfloor }{2^{2lfloor log_2n rfloor}}
$$
proceed like this.



For $n=1$ we have $lfloor log_2n rfloor = 0$.



For $2 le n < 4$ we have $lfloor log_2n rfloor = 1.qquad$ (two terms)



For $4 le n < 8$ we have $lfloor log_2n rfloor = 2.qquad$ (four terms)



And so on,



For $2^k le n < 2^{k+1}$ we have $lfloor log_2n rfloor = k.qquad$ ($2^k$ terms)



So
$$
sumlimits_{n=1}^{infty}dfrac{lfloor log_2n rfloor }{2^{2lfloor log_2n rfloor}} =
sum_{k=0}^infty 2^k;frac{k}{2^{2k}} = 2
$$






share|cite|improve this answer









$endgroup$









  • 1




    $begingroup$
    This has been incredibly helpful, much simpler than expected. Thank you!
    $endgroup$
    – Acas
    Mar 24 at 12:09














9












9








9





$begingroup$

With the integer part still in there
$$
sumlimits_{n=1}^{infty}dfrac{lfloor log_2n rfloor }{2^{2lfloor log_2n rfloor}}
$$
proceed like this.



For $n=1$ we have $lfloor log_2n rfloor = 0$.



For $2 le n < 4$ we have $lfloor log_2n rfloor = 1.qquad$ (two terms)



For $4 le n < 8$ we have $lfloor log_2n rfloor = 2.qquad$ (four terms)



And so on,



For $2^k le n < 2^{k+1}$ we have $lfloor log_2n rfloor = k.qquad$ ($2^k$ terms)



So
$$
sumlimits_{n=1}^{infty}dfrac{lfloor log_2n rfloor }{2^{2lfloor log_2n rfloor}} =
sum_{k=0}^infty 2^k;frac{k}{2^{2k}} = 2
$$






share|cite|improve this answer









$endgroup$



With the integer part still in there
$$
sumlimits_{n=1}^{infty}dfrac{lfloor log_2n rfloor }{2^{2lfloor log_2n rfloor}}
$$
proceed like this.



For $n=1$ we have $lfloor log_2n rfloor = 0$.



For $2 le n < 4$ we have $lfloor log_2n rfloor = 1.qquad$ (two terms)



For $4 le n < 8$ we have $lfloor log_2n rfloor = 2.qquad$ (four terms)



And so on,



For $2^k le n < 2^{k+1}$ we have $lfloor log_2n rfloor = k.qquad$ ($2^k$ terms)



So
$$
sumlimits_{n=1}^{infty}dfrac{lfloor log_2n rfloor }{2^{2lfloor log_2n rfloor}} =
sum_{k=0}^infty 2^k;frac{k}{2^{2k}} = 2
$$







share|cite|improve this answer












share|cite|improve this answer



share|cite|improve this answer










answered Mar 24 at 11:42









GEdgarGEdgar

63.6k269175




63.6k269175








  • 1




    $begingroup$
    This has been incredibly helpful, much simpler than expected. Thank you!
    $endgroup$
    – Acas
    Mar 24 at 12:09














  • 1




    $begingroup$
    This has been incredibly helpful, much simpler than expected. Thank you!
    $endgroup$
    – Acas
    Mar 24 at 12:09








1




1




$begingroup$
This has been incredibly helpful, much simpler than expected. Thank you!
$endgroup$
– Acas
Mar 24 at 12:09




$begingroup$
This has been incredibly helpful, much simpler than expected. Thank you!
$endgroup$
– Acas
Mar 24 at 12:09


















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