Nature of constant $c$ from...
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
asked Oct 29 '18 at 14:38
user514787
695210
|
show 4 more comments
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
asked Oct 29 '18 at 14:38
user514787
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
|
show 4 more comments
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
asked Oct 29 '18 at 14:38
user514787
695210
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
asked Oct 29 '18 at 14:38
user514787
695210
asked Oct 29 '18 at 14:38
user514787
695210
edited Oct 29 '18 at 16:00
asked Oct 29 '18 at 14:38
user514787
695210
asked Oct 29 '18 at 14:38
user514787
695210
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
|
show 4 more comments
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
|
show 4 more comments
1 Answer
1
active
oldest
votes
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).
answered Nov 22 '18 at 2:41
metamorphy
3,6571621
Thank you very much for helping!
– user514787
Nov 22 '18 at 3:00
add a comment |
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1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
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).
answered Nov 22 '18 at 2:41
metamorphy
3,6571621
Thank you very much for helping!
– user514787
Nov 22 '18 at 3:00
add a comment |
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).
answered Nov 22 '18 at 2:41
metamorphy
3,6571621
Thank you very much for helping!
– user514787
Nov 22 '18 at 3:00
add a comment |
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).
answered Nov 22 '18 at 2:41
metamorphy
3,6571621
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).
answered Nov 22 '18 at 2:41
metamorphy
3,6571621
edited Nov 29 '18 at 2:29
answered Nov 22 '18 at 2:41
metamorphy
3,6571621
answered Nov 22 '18 at 2:41
metamorphy
3,6571621
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
add a comment |
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
add a comment |
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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