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Alternating sum of binomial coefficient times logarithm
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This question relates to a statistics question on CrossValidated.SE where the following summation appeared in an answer:
$$I(n) equiv ln n + sum_k=1^n n choose k (-1)^k ln k quad quad quad quad textfor n in mathbbN.$$
Does anyone know of any simpler form for this expression? (I doubt it, but thought I'd check with you brilliant people.) Are there any useful bounds/approximations/asymptotic results for this expression?
summation logarithms binomial-coefficients
edited Jul 16 at 5:12
joriki
164k10180328
asked Jul 16 at 4:18
Ben
81911
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up vote
0
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This question relates to a statistics question on CrossValidated.SE where the following summation appeared in an answer:
$$I(n) equiv ln n + sum_k=1^n n choose k (-1)^k ln k quad quad quad quad textfor n in mathbbN.$$
Does anyone know of any simpler form for this expression? (I doubt it, but thought I'd check with you brilliant people.) Are there any useful bounds/approximations/asymptotic results for this expression?
summation logarithms binomial-coefficients
edited Jul 16 at 5:12
joriki
164k10180328
asked Jul 16 at 4:18
Ben
81911
This might be helpful -- you can transform with $k=n-j$ and subtract $nlog n$, but then you have the case $n=m$ whereas the asymptotic analysis there assumes $ngt m$. Still, you might be able to adapt the analysis.
– joriki
Jul 16 at 5:11
By the way, I get that question suggested as one of the first when I enter your title in "Ask a question". You must have gotten the same display, no?
– joriki
Jul 16 at 5:14
add a comment |Â
up vote
0
down vote
favorite
up vote
0
down vote
favorite
This question relates to a statistics question on CrossValidated.SE where the following summation appeared in an answer:
$$I(n) equiv ln n + sum_k=1^n n choose k (-1)^k ln k quad quad quad quad textfor n in mathbbN.$$
Does anyone know of any simpler form for this expression? (I doubt it, but thought I'd check with you brilliant people.) Are there any useful bounds/approximations/asymptotic results for this expression?
summation logarithms binomial-coefficients
edited Jul 16 at 5:12
joriki
164k10180328
asked Jul 16 at 4:18
Ben
81911
This question relates to a statistics question on CrossValidated.SE where the following summation appeared in an answer:
$$I(n) equiv ln n + sum_k=1^n n choose k (-1)^k ln k quad quad quad quad textfor n in mathbbN.$$
Does anyone know of any simpler form for this expression? (I doubt it, but thought I'd check with you brilliant people.) Are there any useful bounds/approximations/asymptotic results for this expression?
summation logarithms binomial-coefficients
edited Jul 16 at 5:12
joriki
164k10180328
asked Jul 16 at 4:18
Ben
81911
edited Jul 16 at 5:12
joriki
164k10180328
edited Jul 16 at 5:12
joriki
164k10180328
edited Jul 16 at 5:12
joriki
164k10180328
164k10180328
asked Jul 16 at 4:18
Ben
81911
asked Jul 16 at 4:18
Ben
81911
asked Jul 16 at 4:18
Ben
81911
81911
This might be helpful -- you can transform with $k=n-j$ and subtract $nlog n$, but then you have the case $n=m$ whereas the asymptotic analysis there assumes $ngt m$. Still, you might be able to adapt the analysis.
– joriki
Jul 16 at 5:11
By the way, I get that question suggested as one of the first when I enter your title in "Ask a question". You must have gotten the same display, no?
– joriki
Jul 16 at 5:14
add a comment |Â
This might be helpful -- you can transform with $k=n-j$ and subtract $nlog n$, but then you have the case $n=m$ whereas the asymptotic analysis there assumes $ngt m$. Still, you might be able to adapt the analysis.
– joriki
Jul 16 at 5:11
By the way, I get that question suggested as one of the first when I enter your title in "Ask a question". You must have gotten the same display, no?
– joriki
Jul 16 at 5:14
This might be helpful -- you can transform with $k=n-j$ and subtract $nlog n$, but then you have the case $n=m$ whereas the asymptotic analysis there assumes $ngt m$. Still, you might be able to adapt the analysis.
– joriki
Jul 16 at 5:11
This might be helpful -- you can transform with $k=n-j$ and subtract $nlog n$, but then you have the case $n=m$ whereas the asymptotic analysis there assumes $ngt m$. Still, you might be able to adapt the analysis.
– joriki
Jul 16 at 5:11
By the way, I get that question suggested as one of the first when I enter your title in "Ask a question". You must have gotten the same display, no?
– joriki
Jul 16 at 5:14
By the way, I get that question suggested as one of the first when I enter your title in "Ask a question". You must have gotten the same display, no?
– joriki
Jul 16 at 5:14
add a comment |Â
1 Answer
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The large-$n$ asymptotics of this sum are discussed in my paper Asymptotics of Certain Sums Required in Loop Regularisation. I found that the formula
$$ mint_0^infty e^-y(1-e^-y)^m-1logy , dy = -gamma - sum_k geq 1 (-1)^k-1 binomnk logk $$
was most useful: one can apply some standard asymptotic analysis (essentially an extension of Laplace's Method) to this integral to obtain the asymptotic expansion
$$ sum_k geq 1 (-1)^k binomnk logk sim loglogn-gamma - sum_r=1^infty frac1rGamma^(r)(1) (logn)^-r, $$
where $Gamma^(r)$ is the $r$th derivative of the $Gamma$-function.
It seems unlikely that a closed form exists. Closed forms in terms of polynomials in harmonic numbers do exist for sums of this sort with the logarithm replaced by negative integer powers of $k$, by extending some results of Euler (which is also done in my paper), but there's no way to extend the technique to other cases, at least as far as I know: the power being a negative integer is crucial in that case.
answered Jul 16 at 13:36
Chappers
55k74191
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1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
1
down vote
The large-$n$ asymptotics of this sum are discussed in my paper Asymptotics of Certain Sums Required in Loop Regularisation. I found that the formula
$$ mint_0^infty e^-y(1-e^-y)^m-1logy , dy = -gamma - sum_k geq 1 (-1)^k-1 binomnk logk $$
was most useful: one can apply some standard asymptotic analysis (essentially an extension of Laplace's Method) to this integral to obtain the asymptotic expansion
$$ sum_k geq 1 (-1)^k binomnk logk sim loglogn-gamma - sum_r=1^infty frac1rGamma^(r)(1) (logn)^-r, $$
where $Gamma^(r)$ is the $r$th derivative of the $Gamma$-function.
It seems unlikely that a closed form exists. Closed forms in terms of polynomials in harmonic numbers do exist for sums of this sort with the logarithm replaced by negative integer powers of $k$, by extending some results of Euler (which is also done in my paper), but there's no way to extend the technique to other cases, at least as far as I know: the power being a negative integer is crucial in that case.
answered Jul 16 at 13:36
Chappers
55k74191
add a comment |Â
up vote
1
down vote
The large-$n$ asymptotics of this sum are discussed in my paper Asymptotics of Certain Sums Required in Loop Regularisation. I found that the formula
$$ mint_0^infty e^-y(1-e^-y)^m-1logy , dy = -gamma - sum_k geq 1 (-1)^k-1 binomnk logk $$
was most useful: one can apply some standard asymptotic analysis (essentially an extension of Laplace's Method) to this integral to obtain the asymptotic expansion
$$ sum_k geq 1 (-1)^k binomnk logk sim loglogn-gamma - sum_r=1^infty frac1rGamma^(r)(1) (logn)^-r, $$
where $Gamma^(r)$ is the $r$th derivative of the $Gamma$-function.
It seems unlikely that a closed form exists. Closed forms in terms of polynomials in harmonic numbers do exist for sums of this sort with the logarithm replaced by negative integer powers of $k$, by extending some results of Euler (which is also done in my paper), but there's no way to extend the technique to other cases, at least as far as I know: the power being a negative integer is crucial in that case.
answered Jul 16 at 13:36
Chappers
55k74191
add a comment |Â
up vote
1
down vote
up vote
1
down vote
The large-$n$ asymptotics of this sum are discussed in my paper Asymptotics of Certain Sums Required in Loop Regularisation. I found that the formula
$$ mint_0^infty e^-y(1-e^-y)^m-1logy , dy = -gamma - sum_k geq 1 (-1)^k-1 binomnk logk $$
was most useful: one can apply some standard asymptotic analysis (essentially an extension of Laplace's Method) to this integral to obtain the asymptotic expansion
$$ sum_k geq 1 (-1)^k binomnk logk sim loglogn-gamma - sum_r=1^infty frac1rGamma^(r)(1) (logn)^-r, $$
where $Gamma^(r)$ is the $r$th derivative of the $Gamma$-function.
It seems unlikely that a closed form exists. Closed forms in terms of polynomials in harmonic numbers do exist for sums of this sort with the logarithm replaced by negative integer powers of $k$, by extending some results of Euler (which is also done in my paper), but there's no way to extend the technique to other cases, at least as far as I know: the power being a negative integer is crucial in that case.
answered Jul 16 at 13:36
Chappers
55k74191
The large-$n$ asymptotics of this sum are discussed in my paper Asymptotics of Certain Sums Required in Loop Regularisation. I found that the formula
$$ mint_0^infty e^-y(1-e^-y)^m-1logy , dy = -gamma - sum_k geq 1 (-1)^k-1 binomnk logk $$
was most useful: one can apply some standard asymptotic analysis (essentially an extension of Laplace's Method) to this integral to obtain the asymptotic expansion
$$ sum_k geq 1 (-1)^k binomnk logk sim loglogn-gamma - sum_r=1^infty frac1rGamma^(r)(1) (logn)^-r, $$
where $Gamma^(r)$ is the $r$th derivative of the $Gamma$-function.
It seems unlikely that a closed form exists. Closed forms in terms of polynomials in harmonic numbers do exist for sums of this sort with the logarithm replaced by negative integer powers of $k$, by extending some results of Euler (which is also done in my paper), but there's no way to extend the technique to other cases, at least as far as I know: the power being a negative integer is crucial in that case.
answered Jul 16 at 13:36
Chappers
55k74191
answered Jul 16 at 13:36
Chappers
55k74191
answered Jul 16 at 13:36
Chappers
55k74191
answered Jul 16 at 13:36
Chappers
55k74191
55k74191
add a comment |Â
add a comment |Â
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This might be helpful -- you can transform with $k=n-j$ and subtract $nlog n$, but then you have the case $n=m$ whereas the asymptotic analysis there assumes $ngt m$. Still, you might be able to adapt the analysis.
– joriki
Jul 16 at 5:11
By the way, I get that question suggested as one of the first when I enter your title in "Ask a question". You must have gotten the same display, no?
– joriki
Jul 16 at 5:14