Mixing
asymptotic for li(x)-Ri(x)
Clash Royale CLAN TAG#URR8PPP
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Is it true that $$operatornameli(x)-operatornameRi(x) sim frac12operatornameli(x^1/2) (x to infty),$$
where
$$operatornameRi(x) = sum_n = 1^infty fracmu(n)n operatornameli(x^1/n) = 1 + sum_k = 1^infty frac(log x)^kk cdot k! zeta(k+1)$$
for all $x > 0$? If so, how can one prove the given asymptotic?
Note that
beginalignlabellirieq
lim_x to infty fracoperatornameli(x) - operatornameRi(x)frac12operatornameli(x^1/2) = 1- 2lim_x to infty sum_n = 3^infty fracmu(n)nfracoperatornameli(x^1/n)operatornameli(x^1/2) = 1-2sum_n = 3^infty lim_x to infty fracmu(n)nfracoperatornameli(x^1/n)operatornameli(x^1/2) = 1,
endalign
provided that the given limit can be interchanged with the given sum. However, I am unable to justify interchanging the limit with the sum.
nt.number-theory real-analysis ca.classical-analysis-and-odes analytic-number-theory prime-number-theorem
edited Aug 19 at 5:35
GH from MO
56k5136213
asked Aug 18 at 23:24
Jesse Elliott
1,309818
add a comment |Â
up vote
6
down vote
favorite
Is it true that $$operatornameli(x)-operatornameRi(x) sim frac12operatornameli(x^1/2) (x to infty),$$
where
$$operatornameRi(x) = sum_n = 1^infty fracmu(n)n operatornameli(x^1/n) = 1 + sum_k = 1^infty frac(log x)^kk cdot k! zeta(k+1)$$
for all $x > 0$? If so, how can one prove the given asymptotic?
Note that
beginalignlabellirieq
lim_x to infty fracoperatornameli(x) - operatornameRi(x)frac12operatornameli(x^1/2) = 1- 2lim_x to infty sum_n = 3^infty fracmu(n)nfracoperatornameli(x^1/n)operatornameli(x^1/2) = 1-2sum_n = 3^infty lim_x to infty fracmu(n)nfracoperatornameli(x^1/n)operatornameli(x^1/2) = 1,
endalign
provided that the given limit can be interchanged with the given sum. However, I am unable to justify interchanging the limit with the sum.
nt.number-theory real-analysis ca.classical-analysis-and-odes analytic-number-theory prime-number-theorem
edited Aug 19 at 5:35
GH from MO
56k5136213
asked Aug 18 at 23:24
Jesse Elliott
1,309818
add a comment |Â
up vote
6
down vote
favorite
up vote
6
down vote
favorite
Is it true that $$operatornameli(x)-operatornameRi(x) sim frac12operatornameli(x^1/2) (x to infty),$$
where
$$operatornameRi(x) = sum_n = 1^infty fracmu(n)n operatornameli(x^1/n) = 1 + sum_k = 1^infty frac(log x)^kk cdot k! zeta(k+1)$$
for all $x > 0$? If so, how can one prove the given asymptotic?
Note that
beginalignlabellirieq
lim_x to infty fracoperatornameli(x) - operatornameRi(x)frac12operatornameli(x^1/2) = 1- 2lim_x to infty sum_n = 3^infty fracmu(n)nfracoperatornameli(x^1/n)operatornameli(x^1/2) = 1-2sum_n = 3^infty lim_x to infty fracmu(n)nfracoperatornameli(x^1/n)operatornameli(x^1/2) = 1,
endalign
provided that the given limit can be interchanged with the given sum. However, I am unable to justify interchanging the limit with the sum.
nt.number-theory real-analysis ca.classical-analysis-and-odes analytic-number-theory prime-number-theorem
edited Aug 19 at 5:35
GH from MO
56k5136213
asked Aug 18 at 23:24
Jesse Elliott
1,309818
Is it true that $$operatornameli(x)-operatornameRi(x) sim frac12operatornameli(x^1/2) (x to infty),$$
where
$$operatornameRi(x) = sum_n = 1^infty fracmu(n)n operatornameli(x^1/n) = 1 + sum_k = 1^infty frac(log x)^kk cdot k! zeta(k+1)$$
for all $x > 0$? If so, how can one prove the given asymptotic?
Note that
beginalignlabellirieq
lim_x to infty fracoperatornameli(x) - operatornameRi(x)frac12operatornameli(x^1/2) = 1- 2lim_x to infty sum_n = 3^infty fracmu(n)nfracoperatornameli(x^1/n)operatornameli(x^1/2) = 1-2sum_n = 3^infty lim_x to infty fracmu(n)nfracoperatornameli(x^1/n)operatornameli(x^1/2) = 1,
endalign
provided that the given limit can be interchanged with the given sum. However, I am unable to justify interchanging the limit with the sum.
nt.number-theory real-analysis ca.classical-analysis-and-odes analytic-number-theory prime-number-theorem
edited Aug 19 at 5:35
GH from MO
56k5136213
asked Aug 18 at 23:24
Jesse Elliott
1,309818
edited Aug 19 at 5:35
GH from MO
56k5136213
edited Aug 19 at 5:35
GH from MO
56k5136213
edited Aug 19 at 5:35
GH from MO
56k5136213
56k5136213
asked Aug 18 at 23:24
Jesse Elliott
1,309818
asked Aug 18 at 23:24
Jesse Elliott
1,309818
asked Aug 18 at 23:24
Jesse Elliott
1,309818
1,309818
add a comment |Â
add a comment |Â
2 Answers
2
active
oldest
votes
up vote
6
down vote
accepted
Yes, the stated asymptotics (and much more) is true. The idea is to truncate $operatornameRi(x)$ appropriately.
Let us use the series representation (see here)
$$operatornameli(t)=gamma+loglog t+sum_k=1^inftyfrac(log t)^kkcdot k!,qquad t>1.$$
This implies
$$operatornameli(t)=gamma+loglog t+O(log t),qquad 1<t<e,$$
hence also
$$operatornameli(x^1/n)=gamma+loglog x-log n+Oleft(fraclog xnright),qquad n>log x.$$
As a result, for $x>3$ we get
beginalign*sum_n>log x fracmu(n)n operatornameli(x^1/n)&=O(1)+sum_n>log xfracmu(n)nleft(gamma+loglog x-log nright)\[6pt]
&=Oleft((loglog x)^2right)+sum_n=1^inftyfracmu(n)nleft(gamma+loglog x-log nright)\[6pt]
&=Oleft((loglog x)^2right).endalign*
Here we used that
$$sum_n=1^inftyfracmu(n)n=0qquadtextandqquadsum_n=1^inftyfracmu(n)log nn=-1$$
by the prime number theorem. To summarize so far,
$$operatornameRi(x)=sum_nleqlog x fracmu(n)n operatornameli(x^1/n)+Oleft((loglog x)^2right),qquad x>3.$$
The error term on the right hand side can probably be improved, but this is not important. In the sum, we detach the first three terms (corresponding to $n=1,2,3$) and estimate the rest trivially. We get
$$operatornameRi(x)=operatornameli(x)-frac12operatornameli(x^1/2)+Oleft(fracx^1/3log xright),qquad x>3.$$
answered Aug 19 at 5:21
GH from MO
56k5136213
1
Thanks! Your proof yields the stronger fact that the defining series expansion for $operatornameRi(x)$ is also an asymptotic expansion for $operatornameRi(x)$ at $infty$. I suspected this to be true and am glad that your proof confirms it. Strangely I couldn't find a proof of this anywhere in the literature.
– Jesse Elliott
Aug 19 at 21:07
Thanks for accepting my answer, I am glad I could help!
– GH from MO
Aug 19 at 21:59
I'd like to provide this proof in a paper of mine. For now I've cited "GH from MO" and this link. If there is an alternative way you'd like me to cite this you can email me at Jesse.elliott@csuci.edu.
– Jesse Elliott
Aug 21 at 23:39
@JesseElliott: I sent you an email.
– GH from MO
Aug 22 at 12:42
add a comment |Â
up vote
5
down vote
Since we have
$$rm Ri(x)=1+sum_kge 1frac(log x)^kkcdot k!zeta(k+1),$$
which is convergence absolutely for all $x>0$. Therefore we can use the L'Hospital's rule, and by
$$sum_nge 1fracmu(n)n=0$$
we have
beginalign
lim_xrightarrow +inftyfracrm li(x)-rm Ri(x)frac12rm li(x^1/2)&=lim_xrightarrow +inftyfracfrac1log x-frac1xlog xsum_kge 1frac(log x)^kk!zeta(k+1)frac12frac1log(x^1/2)cdotfrac12x^-1/2\
&=lim_xrightarrow +inftyleft(2sqrtxleft(1-frac1xsum_kge 1frac(log x)^kk!sum_nge 1fracmu(n)n^k+1right)right)\
&=lim_xrightarrow +inftyleft(2sqrtxleft(1-frac1xsum_nge 1fracmu(n)n(x^1/n-1)right)right)\
&=1-2lim_xrightarrow +inftyleft(frac1sqrtxsum_nge 3fracmu(n)n(x^1/n-1)right).
endalign
On the other hand, for $x$ sufficiently large,
$$sum_3le nle xfracmu(n)n(x^1/n-1)ll x^1/3log x$$
and
$$sum_n>xfracmu(n)n(x^1/n-1)ll sum_n>xfrac1nleft(expleft(fraclog xnright)-1right)ll sum_n>xfraclog xn^2ll 1.$$
Hence we get
$$lim_xrightarrow +inftyleft(frac1sqrtxsum_nge 3fracmu(n)n(x^1/n-1)right)=0.$$
Which completes the proof.
answered Aug 19 at 2:19
Zhou
33315
add a comment |Â
2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
6
down vote
accepted
Yes, the stated asymptotics (and much more) is true. The idea is to truncate $operatornameRi(x)$ appropriately.
Let us use the series representation (see here)
$$operatornameli(t)=gamma+loglog t+sum_k=1^inftyfrac(log t)^kkcdot k!,qquad t>1.$$
This implies
$$operatornameli(t)=gamma+loglog t+O(log t),qquad 1<t<e,$$
hence also
$$operatornameli(x^1/n)=gamma+loglog x-log n+Oleft(fraclog xnright),qquad n>log x.$$
As a result, for $x>3$ we get
beginalign*sum_n>log x fracmu(n)n operatornameli(x^1/n)&=O(1)+sum_n>log xfracmu(n)nleft(gamma+loglog x-log nright)\[6pt]
&=Oleft((loglog x)^2right)+sum_n=1^inftyfracmu(n)nleft(gamma+loglog x-log nright)\[6pt]
&=Oleft((loglog x)^2right).endalign*
Here we used that
$$sum_n=1^inftyfracmu(n)n=0qquadtextandqquadsum_n=1^inftyfracmu(n)log nn=-1$$
by the prime number theorem. To summarize so far,
$$operatornameRi(x)=sum_nleqlog x fracmu(n)n operatornameli(x^1/n)+Oleft((loglog x)^2right),qquad x>3.$$
The error term on the right hand side can probably be improved, but this is not important. In the sum, we detach the first three terms (corresponding to $n=1,2,3$) and estimate the rest trivially. We get
$$operatornameRi(x)=operatornameli(x)-frac12operatornameli(x^1/2)+Oleft(fracx^1/3log xright),qquad x>3.$$
answered Aug 19 at 5:21
GH from MO
56k5136213
1
Thanks! Your proof yields the stronger fact that the defining series expansion for $operatornameRi(x)$ is also an asymptotic expansion for $operatornameRi(x)$ at $infty$. I suspected this to be true and am glad that your proof confirms it. Strangely I couldn't find a proof of this anywhere in the literature.
– Jesse Elliott
Aug 19 at 21:07
Thanks for accepting my answer, I am glad I could help!
– GH from MO
Aug 19 at 21:59
I'd like to provide this proof in a paper of mine. For now I've cited "GH from MO" and this link. If there is an alternative way you'd like me to cite this you can email me at Jesse.elliott@csuci.edu.
– Jesse Elliott
Aug 21 at 23:39
@JesseElliott: I sent you an email.
– GH from MO
Aug 22 at 12:42
add a comment |Â
up vote
6
down vote
accepted
Yes, the stated asymptotics (and much more) is true. The idea is to truncate $operatornameRi(x)$ appropriately.
Let us use the series representation (see here)
$$operatornameli(t)=gamma+loglog t+sum_k=1^inftyfrac(log t)^kkcdot k!,qquad t>1.$$
This implies
$$operatornameli(t)=gamma+loglog t+O(log t),qquad 1<t<e,$$
hence also
$$operatornameli(x^1/n)=gamma+loglog x-log n+Oleft(fraclog xnright),qquad n>log x.$$
As a result, for $x>3$ we get
beginalign*sum_n>log x fracmu(n)n operatornameli(x^1/n)&=O(1)+sum_n>log xfracmu(n)nleft(gamma+loglog x-log nright)\[6pt]
&=Oleft((loglog x)^2right)+sum_n=1^inftyfracmu(n)nleft(gamma+loglog x-log nright)\[6pt]
&=Oleft((loglog x)^2right).endalign*
Here we used that
$$sum_n=1^inftyfracmu(n)n=0qquadtextandqquadsum_n=1^inftyfracmu(n)log nn=-1$$
by the prime number theorem. To summarize so far,
$$operatornameRi(x)=sum_nleqlog x fracmu(n)n operatornameli(x^1/n)+Oleft((loglog x)^2right),qquad x>3.$$
The error term on the right hand side can probably be improved, but this is not important. In the sum, we detach the first three terms (corresponding to $n=1,2,3$) and estimate the rest trivially. We get
$$operatornameRi(x)=operatornameli(x)-frac12operatornameli(x^1/2)+Oleft(fracx^1/3log xright),qquad x>3.$$
answered Aug 19 at 5:21
GH from MO
56k5136213
1
Thanks! Your proof yields the stronger fact that the defining series expansion for $operatornameRi(x)$ is also an asymptotic expansion for $operatornameRi(x)$ at $infty$. I suspected this to be true and am glad that your proof confirms it. Strangely I couldn't find a proof of this anywhere in the literature.
– Jesse Elliott
Aug 19 at 21:07
Thanks for accepting my answer, I am glad I could help!
– GH from MO
Aug 19 at 21:59
I'd like to provide this proof in a paper of mine. For now I've cited "GH from MO" and this link. If there is an alternative way you'd like me to cite this you can email me at Jesse.elliott@csuci.edu.
– Jesse Elliott
Aug 21 at 23:39
@JesseElliott: I sent you an email.
– GH from MO
Aug 22 at 12:42
add a comment |Â
up vote
6
down vote
accepted
up vote
6
down vote
accepted
Yes, the stated asymptotics (and much more) is true. The idea is to truncate $operatornameRi(x)$ appropriately.
Let us use the series representation (see here)
$$operatornameli(t)=gamma+loglog t+sum_k=1^inftyfrac(log t)^kkcdot k!,qquad t>1.$$
This implies
$$operatornameli(t)=gamma+loglog t+O(log t),qquad 1<t<e,$$
hence also
$$operatornameli(x^1/n)=gamma+loglog x-log n+Oleft(fraclog xnright),qquad n>log x.$$
As a result, for $x>3$ we get
beginalign*sum_n>log x fracmu(n)n operatornameli(x^1/n)&=O(1)+sum_n>log xfracmu(n)nleft(gamma+loglog x-log nright)\[6pt]
&=Oleft((loglog x)^2right)+sum_n=1^inftyfracmu(n)nleft(gamma+loglog x-log nright)\[6pt]
&=Oleft((loglog x)^2right).endalign*
Here we used that
$$sum_n=1^inftyfracmu(n)n=0qquadtextandqquadsum_n=1^inftyfracmu(n)log nn=-1$$
by the prime number theorem. To summarize so far,
$$operatornameRi(x)=sum_nleqlog x fracmu(n)n operatornameli(x^1/n)+Oleft((loglog x)^2right),qquad x>3.$$
The error term on the right hand side can probably be improved, but this is not important. In the sum, we detach the first three terms (corresponding to $n=1,2,3$) and estimate the rest trivially. We get
$$operatornameRi(x)=operatornameli(x)-frac12operatornameli(x^1/2)+Oleft(fracx^1/3log xright),qquad x>3.$$
answered Aug 19 at 5:21
GH from MO
56k5136213
Yes, the stated asymptotics (and much more) is true. The idea is to truncate $operatornameRi(x)$ appropriately.
Let us use the series representation (see here)
$$operatornameli(t)=gamma+loglog t+sum_k=1^inftyfrac(log t)^kkcdot k!,qquad t>1.$$
This implies
$$operatornameli(t)=gamma+loglog t+O(log t),qquad 1<t<e,$$
hence also
$$operatornameli(x^1/n)=gamma+loglog x-log n+Oleft(fraclog xnright),qquad n>log x.$$
As a result, for $x>3$ we get
beginalign*sum_n>log x fracmu(n)n operatornameli(x^1/n)&=O(1)+sum_n>log xfracmu(n)nleft(gamma+loglog x-log nright)\[6pt]
&=Oleft((loglog x)^2right)+sum_n=1^inftyfracmu(n)nleft(gamma+loglog x-log nright)\[6pt]
&=Oleft((loglog x)^2right).endalign*
Here we used that
$$sum_n=1^inftyfracmu(n)n=0qquadtextandqquadsum_n=1^inftyfracmu(n)log nn=-1$$
by the prime number theorem. To summarize so far,
$$operatornameRi(x)=sum_nleqlog x fracmu(n)n operatornameli(x^1/n)+Oleft((loglog x)^2right),qquad x>3.$$
The error term on the right hand side can probably be improved, but this is not important. In the sum, we detach the first three terms (corresponding to $n=1,2,3$) and estimate the rest trivially. We get
$$operatornameRi(x)=operatornameli(x)-frac12operatornameli(x^1/2)+Oleft(fracx^1/3log xright),qquad x>3.$$
answered Aug 19 at 5:21
GH from MO
56k5136213
edited Aug 19 at 7:20
answered Aug 19 at 5:21
GH from MO
56k5136213
answered Aug 19 at 5:21
GH from MO
56k5136213
answered Aug 19 at 5:21
GH from MO
56k5136213
56k5136213
1
Thanks! Your proof yields the stronger fact that the defining series expansion for $operatornameRi(x)$ is also an asymptotic expansion for $operatornameRi(x)$ at $infty$. I suspected this to be true and am glad that your proof confirms it. Strangely I couldn't find a proof of this anywhere in the literature.
– Jesse Elliott
Aug 19 at 21:07
Thanks for accepting my answer, I am glad I could help!
– GH from MO
Aug 19 at 21:59
I'd like to provide this proof in a paper of mine. For now I've cited "GH from MO" and this link. If there is an alternative way you'd like me to cite this you can email me at Jesse.elliott@csuci.edu.
– Jesse Elliott
Aug 21 at 23:39
@JesseElliott: I sent you an email.
– GH from MO
Aug 22 at 12:42
add a comment |Â
1
Thanks! Your proof yields the stronger fact that the defining series expansion for $operatornameRi(x)$ is also an asymptotic expansion for $operatornameRi(x)$ at $infty$. I suspected this to be true and am glad that your proof confirms it. Strangely I couldn't find a proof of this anywhere in the literature.
– Jesse Elliott
Aug 19 at 21:07
Thanks for accepting my answer, I am glad I could help!
– GH from MO
Aug 19 at 21:59
I'd like to provide this proof in a paper of mine. For now I've cited "GH from MO" and this link. If there is an alternative way you'd like me to cite this you can email me at Jesse.elliott@csuci.edu.
– Jesse Elliott
Aug 21 at 23:39
@JesseElliott: I sent you an email.
– GH from MO
Aug 22 at 12:42
1
1
Thanks! Your proof yields the stronger fact that the defining series expansion for $operatornameRi(x)$ is also an asymptotic expansion for $operatornameRi(x)$ at $infty$. I suspected this to be true and am glad that your proof confirms it. Strangely I couldn't find a proof of this anywhere in the literature.
– Jesse Elliott
Aug 19 at 21:07
Thanks! Your proof yields the stronger fact that the defining series expansion for $operatornameRi(x)$ is also an asymptotic expansion for $operatornameRi(x)$ at $infty$. I suspected this to be true and am glad that your proof confirms it. Strangely I couldn't find a proof of this anywhere in the literature.
– Jesse Elliott
Aug 19 at 21:07
Thanks for accepting my answer, I am glad I could help!
– GH from MO
Aug 19 at 21:59
Thanks for accepting my answer, I am glad I could help!
– GH from MO
Aug 19 at 21:59
I'd like to provide this proof in a paper of mine. For now I've cited "GH from MO" and this link. If there is an alternative way you'd like me to cite this you can email me at Jesse.elliott@csuci.edu.
– Jesse Elliott
Aug 21 at 23:39
I'd like to provide this proof in a paper of mine. For now I've cited "GH from MO" and this link. If there is an alternative way you'd like me to cite this you can email me at Jesse.elliott@csuci.edu.
– Jesse Elliott
Aug 21 at 23:39
@JesseElliott: I sent you an email.
– GH from MO
Aug 22 at 12:42
@JesseElliott: I sent you an email.
– GH from MO
Aug 22 at 12:42
add a comment |Â
up vote
5
down vote
Since we have
$$rm Ri(x)=1+sum_kge 1frac(log x)^kkcdot k!zeta(k+1),$$
which is convergence absolutely for all $x>0$. Therefore we can use the L'Hospital's rule, and by
$$sum_nge 1fracmu(n)n=0$$
we have
beginalign
lim_xrightarrow +inftyfracrm li(x)-rm Ri(x)frac12rm li(x^1/2)&=lim_xrightarrow +inftyfracfrac1log x-frac1xlog xsum_kge 1frac(log x)^kk!zeta(k+1)frac12frac1log(x^1/2)cdotfrac12x^-1/2\
&=lim_xrightarrow +inftyleft(2sqrtxleft(1-frac1xsum_kge 1frac(log x)^kk!sum_nge 1fracmu(n)n^k+1right)right)\
&=lim_xrightarrow +inftyleft(2sqrtxleft(1-frac1xsum_nge 1fracmu(n)n(x^1/n-1)right)right)\
&=1-2lim_xrightarrow +inftyleft(frac1sqrtxsum_nge 3fracmu(n)n(x^1/n-1)right).
endalign
On the other hand, for $x$ sufficiently large,
$$sum_3le nle xfracmu(n)n(x^1/n-1)ll x^1/3log x$$
and
$$sum_n>xfracmu(n)n(x^1/n-1)ll sum_n>xfrac1nleft(expleft(fraclog xnright)-1right)ll sum_n>xfraclog xn^2ll 1.$$
Hence we get
$$lim_xrightarrow +inftyleft(frac1sqrtxsum_nge 3fracmu(n)n(x^1/n-1)right)=0.$$
Which completes the proof.
answered Aug 19 at 2:19
Zhou
33315
add a comment |Â
up vote
5
down vote
Since we have
$$rm Ri(x)=1+sum_kge 1frac(log x)^kkcdot k!zeta(k+1),$$
which is convergence absolutely for all $x>0$. Therefore we can use the L'Hospital's rule, and by
$$sum_nge 1fracmu(n)n=0$$
we have
beginalign
lim_xrightarrow +inftyfracrm li(x)-rm Ri(x)frac12rm li(x^1/2)&=lim_xrightarrow +inftyfracfrac1log x-frac1xlog xsum_kge 1frac(log x)^kk!zeta(k+1)frac12frac1log(x^1/2)cdotfrac12x^-1/2\
&=lim_xrightarrow +inftyleft(2sqrtxleft(1-frac1xsum_kge 1frac(log x)^kk!sum_nge 1fracmu(n)n^k+1right)right)\
&=lim_xrightarrow +inftyleft(2sqrtxleft(1-frac1xsum_nge 1fracmu(n)n(x^1/n-1)right)right)\
&=1-2lim_xrightarrow +inftyleft(frac1sqrtxsum_nge 3fracmu(n)n(x^1/n-1)right).
endalign
On the other hand, for $x$ sufficiently large,
$$sum_3le nle xfracmu(n)n(x^1/n-1)ll x^1/3log x$$
and
$$sum_n>xfracmu(n)n(x^1/n-1)ll sum_n>xfrac1nleft(expleft(fraclog xnright)-1right)ll sum_n>xfraclog xn^2ll 1.$$
Hence we get
$$lim_xrightarrow +inftyleft(frac1sqrtxsum_nge 3fracmu(n)n(x^1/n-1)right)=0.$$
Which completes the proof.
answered Aug 19 at 2:19
Zhou
33315
add a comment |Â
up vote
5
down vote
up vote
5
down vote
Since we have
$$rm Ri(x)=1+sum_kge 1frac(log x)^kkcdot k!zeta(k+1),$$
which is convergence absolutely for all $x>0$. Therefore we can use the L'Hospital's rule, and by
$$sum_nge 1fracmu(n)n=0$$
we have
beginalign
lim_xrightarrow +inftyfracrm li(x)-rm Ri(x)frac12rm li(x^1/2)&=lim_xrightarrow +inftyfracfrac1log x-frac1xlog xsum_kge 1frac(log x)^kk!zeta(k+1)frac12frac1log(x^1/2)cdotfrac12x^-1/2\
&=lim_xrightarrow +inftyleft(2sqrtxleft(1-frac1xsum_kge 1frac(log x)^kk!sum_nge 1fracmu(n)n^k+1right)right)\
&=lim_xrightarrow +inftyleft(2sqrtxleft(1-frac1xsum_nge 1fracmu(n)n(x^1/n-1)right)right)\
&=1-2lim_xrightarrow +inftyleft(frac1sqrtxsum_nge 3fracmu(n)n(x^1/n-1)right).
endalign
On the other hand, for $x$ sufficiently large,
$$sum_3le nle xfracmu(n)n(x^1/n-1)ll x^1/3log x$$
and
$$sum_n>xfracmu(n)n(x^1/n-1)ll sum_n>xfrac1nleft(expleft(fraclog xnright)-1right)ll sum_n>xfraclog xn^2ll 1.$$
Hence we get
$$lim_xrightarrow +inftyleft(frac1sqrtxsum_nge 3fracmu(n)n(x^1/n-1)right)=0.$$
Which completes the proof.
answered Aug 19 at 2:19
Zhou
33315
Since we have
$$rm Ri(x)=1+sum_kge 1frac(log x)^kkcdot k!zeta(k+1),$$
which is convergence absolutely for all $x>0$. Therefore we can use the L'Hospital's rule, and by
$$sum_nge 1fracmu(n)n=0$$
we have
beginalign
lim_xrightarrow +inftyfracrm li(x)-rm Ri(x)frac12rm li(x^1/2)&=lim_xrightarrow +inftyfracfrac1log x-frac1xlog xsum_kge 1frac(log x)^kk!zeta(k+1)frac12frac1log(x^1/2)cdotfrac12x^-1/2\
&=lim_xrightarrow +inftyleft(2sqrtxleft(1-frac1xsum_kge 1frac(log x)^kk!sum_nge 1fracmu(n)n^k+1right)right)\
&=lim_xrightarrow +inftyleft(2sqrtxleft(1-frac1xsum_nge 1fracmu(n)n(x^1/n-1)right)right)\
&=1-2lim_xrightarrow +inftyleft(frac1sqrtxsum_nge 3fracmu(n)n(x^1/n-1)right).
endalign
On the other hand, for $x$ sufficiently large,
$$sum_3le nle xfracmu(n)n(x^1/n-1)ll x^1/3log x$$
and
$$sum_n>xfracmu(n)n(x^1/n-1)ll sum_n>xfrac1nleft(expleft(fraclog xnright)-1right)ll sum_n>xfraclog xn^2ll 1.$$
Hence we get
$$lim_xrightarrow +inftyleft(frac1sqrtxsum_nge 3fracmu(n)n(x^1/n-1)right)=0.$$
Which completes the proof.
answered Aug 19 at 2:19
Zhou
33315
answered Aug 19 at 2:19
Zhou
33315
answered Aug 19 at 2:19
Zhou
33315
answered Aug 19 at 2:19
Zhou
33315
33315
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