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 at 14:46










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






  • 1




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










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






  • 1




    I get $c=0.00900716488816377650177841067803580738540696574162ldots$
    – metamorphy
    Nov 22 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 at 14:46










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






  • 1




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










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






  • 1




    I get $c=0.00900716488816377650177841067803580738540696574162ldots$
    – metamorphy
    Nov 22 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 at 16:00

























asked Oct 29 at 14:38









user514787

683210




683210












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










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






  • 1




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










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






  • 1




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


















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










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






  • 1




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










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






  • 1




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
















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




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












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




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




1




1




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




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












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




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




1




1




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




I get $c=0.00900716488816377650177841067803580738540696574162ldots$
– metamorphy
Nov 22 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 at 3:00











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1 Answer
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1 Answer
1






active

oldest

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active

oldest

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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 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 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 at 2:29

























answered Nov 22 at 2:41









metamorphy

3,2221520




3,2221520












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


















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
















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




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


















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