Nature of constant $c$ from...












2














If we take
$$sumlimits_{k=1}^{n}(1/k)^{1/k}=a(n)$$
so
$$limlimits_{ntoinfty}left(a(n)-n+frac{log^2(n)}{2}-1right)=c$$
What is the nature of constant $c$? Is it really constant (maybe it function or sum of function and constant)?






summation logarithms approximation constants







share|cite|improve this question
























  • The RHS makes no sense, it contains $n$ while the limit on the LHS does not. Please correct.
    – gammatester
    Oct 29 '18 at 14:46










  • @gammatester, thank you for comment! Better now?
    – user514787
    Oct 29 '18 at 14:54






  • 1




    Is my edit correct?
    – gammatester
    Oct 29 '18 at 14:58










  • @gammatester, absolutely.
    – user514787
    Oct 29 '18 at 15:00






  • 1




    I get $c=0.00900716488816377650177841067803580738540696574162ldots$
    – metamorphy
    Nov 22 '18 at 1:34
















2














If we take
$$sumlimits_{k=1}^{n}(1/k)^{1/k}=a(n)$$
so
$$limlimits_{ntoinfty}left(a(n)-n+frac{log^2(n)}{2}-1right)=c$$
What is the nature of constant $c$? Is it really constant (maybe it function or sum of function and constant)?






summation logarithms approximation constants







share|cite|improve this question
























  • The RHS makes no sense, it contains $n$ while the limit on the LHS does not. Please correct.
    – gammatester
    Oct 29 '18 at 14:46










  • @gammatester, thank you for comment! Better now?
    – user514787
    Oct 29 '18 at 14:54






  • 1




    Is my edit correct?
    – gammatester
    Oct 29 '18 at 14:58










  • @gammatester, absolutely.
    – user514787
    Oct 29 '18 at 15:00






  • 1




    I get $c=0.00900716488816377650177841067803580738540696574162ldots$
    – metamorphy
    Nov 22 '18 at 1:34














2












2








2


1





If we take
$$sumlimits_{k=1}^{n}(1/k)^{1/k}=a(n)$$
so
$$limlimits_{ntoinfty}left(a(n)-n+frac{log^2(n)}{2}-1right)=c$$
What is the nature of constant $c$? Is it really constant (maybe it function or sum of function and constant)?






summation logarithms approximation constants







share|cite|improve this question















If we take
$$sumlimits_{k=1}^{n}(1/k)^{1/k}=a(n)$$
so
$$limlimits_{ntoinfty}left(a(n)-n+frac{log^2(n)}{2}-1right)=c$$
What is the nature of constant $c$? Is it really constant (maybe it function or sum of function and constant)?






summation logarithms approximation constants




summation logarithms approximation constants






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Oct 29 '18 at 16:00

























asked Oct 29 '18 at 14:38









user514787

695210




695210












  • The RHS makes no sense, it contains $n$ while the limit on the LHS does not. Please correct.
    – gammatester
    Oct 29 '18 at 14:46










  • @gammatester, thank you for comment! Better now?
    – user514787
    Oct 29 '18 at 14:54






  • 1




    Is my edit correct?
    – gammatester
    Oct 29 '18 at 14:58










  • @gammatester, absolutely.
    – user514787
    Oct 29 '18 at 15:00






  • 1




    I get $c=0.00900716488816377650177841067803580738540696574162ldots$
    – metamorphy
    Nov 22 '18 at 1:34


















  • The RHS makes no sense, it contains $n$ while the limit on the LHS does not. Please correct.
    – gammatester
    Oct 29 '18 at 14:46










  • @gammatester, thank you for comment! Better now?
    – user514787
    Oct 29 '18 at 14:54






  • 1




    Is my edit correct?
    – gammatester
    Oct 29 '18 at 14:58










  • @gammatester, absolutely.
    – user514787
    Oct 29 '18 at 15:00






  • 1




    I get $c=0.00900716488816377650177841067803580738540696574162ldots$
    – metamorphy
    Nov 22 '18 at 1:34
















The RHS makes no sense, it contains $n$ while the limit on the LHS does not. Please correct.
– gammatester
Oct 29 '18 at 14:46




The RHS makes no sense, it contains $n$ while the limit on the LHS does not. Please correct.
– gammatester
Oct 29 '18 at 14:46












@gammatester, thank you for comment! Better now?
– user514787
Oct 29 '18 at 14:54




@gammatester, thank you for comment! Better now?
– user514787
Oct 29 '18 at 14:54




1




1




Is my edit correct?
– gammatester
Oct 29 '18 at 14:58




Is my edit correct?
– gammatester
Oct 29 '18 at 14:58












@gammatester, absolutely.
– user514787
Oct 29 '18 at 15:00




@gammatester, absolutely.
– user514787
Oct 29 '18 at 15:00




1




1




I get $c=0.00900716488816377650177841067803580738540696574162ldots$
– metamorphy
Nov 22 '18 at 1:34




I get $c=0.00900716488816377650177841067803580738540696574162ldots$
– metamorphy
Nov 22 '18 at 1:34










1 Answer
1






active

oldest

votes


















1














Computed it using PARI/GP (this is not an answer to the question on "nature"...). Write
$$1+c=sum_{n=1}^{infty}left((1/n)^{1/n}-1+frac{ln^2(n+1)-ln^2n}{2}right)=F+G$$
(PARI fails to compute it in this form), where $F=displaystylesum_{n=1}^{infty}f(n)$, $G=displaystylesum_{n=1}^{infty}gBig(frac{ln n}{n}Big)$,
$$f(z)=frac{ln^2(1+z)-ln^2z}{2}-frac{ln z}{z},quad g(z)=e^{-z}-1+z.$$
Here, PARI's $texttt{sumnum}$ and $texttt{sumnumap}$ compute $G$ correctly, but not $F$. It seems that in
$$F=frac{f(1)}{2}+int_1^infty f(x),dx+iint_0^inftyfrac{f(1+it)-f(1-it)}{e^{2pi t}-1},dt$$
(this is Abel-Plana formula), PARI cannot (or I couldn't make it) compute the elementary
$$int_1^infty f(x),dx=-(1-ln2)^2.$$
But, armed with these handmade interventions, I get it working:



f(z)=(log(1+z)^2-log(z)^2)/2-log(z)/z;

g(t)=imag(f(1-I*t)-f(1+I*t))/(exp(2*Pi*t)-1);

a(n)={my(r=log(n)/n);return(exp(-r)-1+r)};

b=(log(2)/2)^2-(1-log(2))^2-1;

b+sumnum(n=1,a(n))+intnum(t=0,[+oo,2*Pi],g(t))


gives $0.0090071648881637765017784106780358074$ (with the default precision).






share|cite|improve this answer























  • Thank you very much for helping!
    – user514787
    Nov 22 '18 at 3:00











Your Answer





StackExchange.ifUsing("editor", function () {
return StackExchange.using("mathjaxEditing", function () {
StackExchange.MarkdownEditor.creationCallbacks.add(function (editor, postfix) {
StackExchange.mathjaxEditing.prepareWmdForMathJax(editor, postfix, [["$", "$"], ["\\(","\\)"]]);
});
});
}, "mathjax-editing");

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
});


}
});














draft saved

draft discarded


















StackExchange.ready(
function () {
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f2976201%2fnature-of-constant-c-from-lim-limits-n-to-infty-left-sum-limits-k-1n%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









1














Computed it using PARI/GP (this is not an answer to the question on "nature"...). Write
$$1+c=sum_{n=1}^{infty}left((1/n)^{1/n}-1+frac{ln^2(n+1)-ln^2n}{2}right)=F+G$$
(PARI fails to compute it in this form), where $F=displaystylesum_{n=1}^{infty}f(n)$, $G=displaystylesum_{n=1}^{infty}gBig(frac{ln n}{n}Big)$,
$$f(z)=frac{ln^2(1+z)-ln^2z}{2}-frac{ln z}{z},quad g(z)=e^{-z}-1+z.$$
Here, PARI's $texttt{sumnum}$ and $texttt{sumnumap}$ compute $G$ correctly, but not $F$. It seems that in
$$F=frac{f(1)}{2}+int_1^infty f(x),dx+iint_0^inftyfrac{f(1+it)-f(1-it)}{e^{2pi t}-1},dt$$
(this is Abel-Plana formula), PARI cannot (or I couldn't make it) compute the elementary
$$int_1^infty f(x),dx=-(1-ln2)^2.$$
But, armed with these handmade interventions, I get it working:



f(z)=(log(1+z)^2-log(z)^2)/2-log(z)/z;

g(t)=imag(f(1-I*t)-f(1+I*t))/(exp(2*Pi*t)-1);

a(n)={my(r=log(n)/n);return(exp(-r)-1+r)};

b=(log(2)/2)^2-(1-log(2))^2-1;

b+sumnum(n=1,a(n))+intnum(t=0,[+oo,2*Pi],g(t))


gives $0.0090071648881637765017784106780358074$ (with the default precision).






share|cite|improve this answer























  • Thank you very much for helping!
    – user514787
    Nov 22 '18 at 3:00
















1














Computed it using PARI/GP (this is not an answer to the question on "nature"...). Write
$$1+c=sum_{n=1}^{infty}left((1/n)^{1/n}-1+frac{ln^2(n+1)-ln^2n}{2}right)=F+G$$
(PARI fails to compute it in this form), where $F=displaystylesum_{n=1}^{infty}f(n)$, $G=displaystylesum_{n=1}^{infty}gBig(frac{ln n}{n}Big)$,
$$f(z)=frac{ln^2(1+z)-ln^2z}{2}-frac{ln z}{z},quad g(z)=e^{-z}-1+z.$$
Here, PARI's $texttt{sumnum}$ and $texttt{sumnumap}$ compute $G$ correctly, but not $F$. It seems that in
$$F=frac{f(1)}{2}+int_1^infty f(x),dx+iint_0^inftyfrac{f(1+it)-f(1-it)}{e^{2pi t}-1},dt$$
(this is Abel-Plana formula), PARI cannot (or I couldn't make it) compute the elementary
$$int_1^infty f(x),dx=-(1-ln2)^2.$$
But, armed with these handmade interventions, I get it working:



f(z)=(log(1+z)^2-log(z)^2)/2-log(z)/z;

g(t)=imag(f(1-I*t)-f(1+I*t))/(exp(2*Pi*t)-1);

a(n)={my(r=log(n)/n);return(exp(-r)-1+r)};

b=(log(2)/2)^2-(1-log(2))^2-1;

b+sumnum(n=1,a(n))+intnum(t=0,[+oo,2*Pi],g(t))


gives $0.0090071648881637765017784106780358074$ (with the default precision).






share|cite|improve this answer























  • Thank you very much for helping!
    – user514787
    Nov 22 '18 at 3:00














1












1








1






Computed it using PARI/GP (this is not an answer to the question on "nature"...). Write
$$1+c=sum_{n=1}^{infty}left((1/n)^{1/n}-1+frac{ln^2(n+1)-ln^2n}{2}right)=F+G$$
(PARI fails to compute it in this form), where $F=displaystylesum_{n=1}^{infty}f(n)$, $G=displaystylesum_{n=1}^{infty}gBig(frac{ln n}{n}Big)$,
$$f(z)=frac{ln^2(1+z)-ln^2z}{2}-frac{ln z}{z},quad g(z)=e^{-z}-1+z.$$
Here, PARI's $texttt{sumnum}$ and $texttt{sumnumap}$ compute $G$ correctly, but not $F$. It seems that in
$$F=frac{f(1)}{2}+int_1^infty f(x),dx+iint_0^inftyfrac{f(1+it)-f(1-it)}{e^{2pi t}-1},dt$$
(this is Abel-Plana formula), PARI cannot (or I couldn't make it) compute the elementary
$$int_1^infty f(x),dx=-(1-ln2)^2.$$
But, armed with these handmade interventions, I get it working:



f(z)=(log(1+z)^2-log(z)^2)/2-log(z)/z;

g(t)=imag(f(1-I*t)-f(1+I*t))/(exp(2*Pi*t)-1);

a(n)={my(r=log(n)/n);return(exp(-r)-1+r)};

b=(log(2)/2)^2-(1-log(2))^2-1;

b+sumnum(n=1,a(n))+intnum(t=0,[+oo,2*Pi],g(t))


gives $0.0090071648881637765017784106780358074$ (with the default precision).






share|cite|improve this answer














Computed it using PARI/GP (this is not an answer to the question on "nature"...). Write
$$1+c=sum_{n=1}^{infty}left((1/n)^{1/n}-1+frac{ln^2(n+1)-ln^2n}{2}right)=F+G$$
(PARI fails to compute it in this form), where $F=displaystylesum_{n=1}^{infty}f(n)$, $G=displaystylesum_{n=1}^{infty}gBig(frac{ln n}{n}Big)$,
$$f(z)=frac{ln^2(1+z)-ln^2z}{2}-frac{ln z}{z},quad g(z)=e^{-z}-1+z.$$
Here, PARI's $texttt{sumnum}$ and $texttt{sumnumap}$ compute $G$ correctly, but not $F$. It seems that in
$$F=frac{f(1)}{2}+int_1^infty f(x),dx+iint_0^inftyfrac{f(1+it)-f(1-it)}{e^{2pi t}-1},dt$$
(this is Abel-Plana formula), PARI cannot (or I couldn't make it) compute the elementary
$$int_1^infty f(x),dx=-(1-ln2)^2.$$
But, armed with these handmade interventions, I get it working:



f(z)=(log(1+z)^2-log(z)^2)/2-log(z)/z;

g(t)=imag(f(1-I*t)-f(1+I*t))/(exp(2*Pi*t)-1);

a(n)={my(r=log(n)/n);return(exp(-r)-1+r)};

b=(log(2)/2)^2-(1-log(2))^2-1;

b+sumnum(n=1,a(n))+intnum(t=0,[+oo,2*Pi],g(t))


gives $0.0090071648881637765017784106780358074$ (with the default precision).







share|cite|improve this answer














share|cite|improve this answer



share|cite|improve this answer








edited Nov 29 '18 at 2:29

























answered Nov 22 '18 at 2:41









metamorphy

3,6571621




3,6571621












  • Thank you very much for helping!
    – user514787
    Nov 22 '18 at 3:00


















  • Thank you very much for helping!
    – user514787
    Nov 22 '18 at 3:00
















Thank you very much for helping!
– user514787
Nov 22 '18 at 3:00




Thank you very much for helping!
– user514787
Nov 22 '18 at 3:00


















draft saved

draft discarded




















































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.





Some of your past answers have not been well-received, and you're in danger of being blocked from answering.


Please pay close attention to the following guidance:


  • 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.


To learn more, see our tips on writing great answers.




draft saved


draft discarded














StackExchange.ready(
function () {
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f2976201%2fnature-of-constant-c-from-lim-limits-n-to-infty-left-sum-limits-k-1n%23new-answer', 'question_page');
}
);

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







-

Popular posts from this blog

mysqli_query(): Empty query in /home/lucindabrummitt/public_html/blog/wp-includes/wp-db.php on line 1924

Image
0 I just uploaded a new plugin to my wordpress site and I got this warning: mysqli_query(): Empty query in /home/lucindabrummitt/public_html/blog/wp-includes/wp-db.php on line 1924 Usually if I have a problem, I can find answers on google but this one seems pretty particular and I'm not finding any info on it except for this forum. I'm not terribly techie but if someone can tell me how to read this and how to change it I can follow directions pretty well. Thanks All! PS If I don't have the proper tags here could someone add the appropriate tags? php mysql wordpress share | improve this question asked Nov 16 '18 at 20:41 ...
Read more

How to change which sound is reproduced for terminal bell?

Image
3 Recently upgrade to 18.10 from 18.04, and using the terminal is becoming annoying: the new alert sound is a drip ( /usr/share/sounds/gnome/default/alerts/drip.ogg ), and the previous one is still located at ( /usr/share/sounds/ubuntu/stereo/bell.ogg ). My issue is that at the gnome-terminal, when autocompleting or pressing left BOTH sounds are reproduced, at the same time. I want to just have ubuntu/stereo/bell.ogg active, not both . At the moment I have been only able to reduce the volume of the alerts (but still both are played), or turn the alerts down (but I dislike both "solutions"). gnome-terminal 18.10 teminal-bell share | improve this question ...
Read more

Can I use Tabulator js library in my java Spring + Thymeleaf project?

Image
0 I want to add Tabulator js to my java project. I added tabulator.min.js, tabulator.min.css to resources folder. After that added to my html page these dependencies as recommending in readme.md. <link rel="stylesheet" th:href="@{/css/tabulator.min.css}"> <script type="text/javascript" th:src="@{/js/tabulator.min.js}"></script> Now my tables view all elements, that I have in the html page, but not correct. In the documentation of Tabulator is example how to create header and table, var table = new Tabulator("#example-table", { columns:[ {title:"Name", field:"name", sorter:"string", width:200, editor:true}, {title:"Age", field:"age", sorter:"...
Read more