Is it known the value of $sumlimits_{n=1}^{infty}frac{log(n)}{ n^2}$? Announcing the arrival...
Why are there no cargo aircraft with "flying wing" design?
Identify plant with long narrow paired leaves and reddish stems
How to run gsettings for another user Ubuntu 18.04.2 LTS
Book where humans were engineered with genes from animal species to survive hostile planets
When a candle burns, why does the top of wick glow if bottom of flame is hottest?
What's the meaning of 間時肆拾貳 at a car parking sign
How to find all the available tools in mac terminal?
Generate an RGB colour grid
Is it ethical to give a final exam after the professor has quit before teaching the remaining chapters of the course?
Storing hydrofluoric acid before the invention of plastics
String `!23` is replaced with `docker` in command line
What's the purpose of writing one's academic biography in the third person?
How does the particle を relate to the verb 行く in the structure「A を + B に行く」?
Is it fair for a professor to grade us on the possession of past papers?
If a contract sometimes uses the wrong name, is it still valid?
Sci-Fi book where patients in a coma ward all live in a subconscious world linked together
Apollo command module space walk?
Fundamental Solution of the Pell Equation
What is Arya's weapon design?
Why light coming from distant stars is not discreet?
Output the ŋarâþ crîþ alphabet song without using (m)any letters
Error "illegal generic type for instanceof" when using local classes
What is the meaning of the new sigil in Game of Thrones Season 8 intro?
Dating a Former Employee
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.
$begingroup$
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
$endgroup$
add a comment |
$begingroup$
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
$endgroup$
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
add a comment |
$begingroup$
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
$endgroup$
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
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
add a comment |
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
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
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
$$
sumlimits_{n=1}^{infty}dfrac{lfloor log_2n rfloor }{2^{2lfloor log_2n rfloor}} =
sum_{k=0}^infty 2^k;frac{k}{2^{2k}} = 2
$$
$endgroup$
1
$begingroup$
This has been incredibly helpful, much simpler than expected. Thank you!
$endgroup$
– Acas
Mar 24 at 12:09
add a comment |
Your Answer
StackExchange.ready(function() {
var channelOptions = {
tags: "".split(" "),
id: "69"
};
initTagRenderer("".split(" "), "".split(" "), channelOptions);
StackExchange.using("externalEditor", function() {
// Have to fire editor after snippets, if snippets enabled
if (StackExchange.settings.snippets.snippetsEnabled) {
StackExchange.using("snippets", function() {
createEditor();
});
}
else {
createEditor();
}
});
function createEditor() {
StackExchange.prepareEditor({
heartbeatType: 'answer',
autoActivateHeartbeat: false,
convertImagesToLinks: true,
noModals: true,
showLowRepImageUploadWarning: true,
reputationToPostImages: 10,
bindNavPrevention: true,
postfix: "",
imageUploader: {
brandingHtml: "Powered by u003ca class="icon-imgur-white" href="https://imgur.com/"u003eu003c/au003e",
contentPolicyHtml: "User contributions licensed under u003ca href="https://creativecommons.org/licenses/by-sa/3.0/"u003ecc by-sa 3.0 with attribution requiredu003c/au003e u003ca href="https://stackoverflow.com/legal/content-policy"u003e(content policy)u003c/au003e",
allowUrls: true
},
noCode: true, onDemand: true,
discardSelector: ".discard-answer"
,immediatelyShowMarkdownHelp:true
});
}
});
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
StackExchange.ready(
function () {
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3160410%2fis-it-known-the-value-of-sum-limits-n-1-infty-frac-logn-n2%23new-answer', 'question_page');
}
);
Post as a guest
Required, but never shown
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
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
$$
sumlimits_{n=1}^{infty}dfrac{lfloor log_2n rfloor }{2^{2lfloor log_2n rfloor}} =
sum_{k=0}^infty 2^k;frac{k}{2^{2k}} = 2
$$
$endgroup$
1
$begingroup$
This has been incredibly helpful, much simpler than expected. Thank you!
$endgroup$
– Acas
Mar 24 at 12:09
add a comment |
$begingroup$
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
$$
sumlimits_{n=1}^{infty}dfrac{lfloor log_2n rfloor }{2^{2lfloor log_2n rfloor}} =
sum_{k=0}^infty 2^k;frac{k}{2^{2k}} = 2
$$
$endgroup$
1
$begingroup$
This has been incredibly helpful, much simpler than expected. Thank you!
$endgroup$
– Acas
Mar 24 at 12:09
add a comment |
$begingroup$
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
$$
sumlimits_{n=1}^{infty}dfrac{lfloor log_2n rfloor }{2^{2lfloor log_2n rfloor}} =
sum_{k=0}^infty 2^k;frac{k}{2^{2k}} = 2
$$
$endgroup$
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
$$
sumlimits_{n=1}^{infty}dfrac{lfloor log_2n rfloor }{2^{2lfloor log_2n rfloor}} =
sum_{k=0}^infty 2^k;frac{k}{2^{2k}} = 2
$$
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
add a comment |
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
add a comment |
Thanks for contributing an answer to Mathematics Stack Exchange!
- Please be sure to answer the question. Provide details and share your research!
But avoid …
- Asking for help, clarification, or responding to other answers.
- Making statements based on opinion; back them up with references or personal experience.
Use MathJax to format equations. MathJax reference.
To learn more, see our tips on writing great answers.
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
StackExchange.ready(
function () {
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3160410%2fis-it-known-the-value-of-sum-limits-n-1-infty-frac-logn-n2%23new-answer', 'question_page');
}
);
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
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