Logarithm as a limit of sums

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Trying to understand a paper which makes use of the following "fact" about logarithms. Can someone please explain why it is so?



Let $phi(x) = sum_x/a leq nleq x/b 1/n$. Then $lim_xto inftyphi(x) = log(a/b)$. In other words,
$$log(a/b) = lim_xto inftysum_x/a leq nleq x/bfrac1n.$$




analysis logarithms




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    Do you know/can you show that $$sum_n leqslant y frac1n = log y + gamma + Obiggl(frac1ybiggr),?$$
    – Daniel Fischer♦
    Jul 16 at 13:17














up vote
2
down vote

favorite












Trying to understand a paper which makes use of the following "fact" about logarithms. Can someone please explain why it is so?



Let $phi(x) = sum_x/a leq nleq x/b 1/n$. Then $lim_xto inftyphi(x) = log(a/b)$. In other words,
$$log(a/b) = lim_xto inftysum_x/a leq nleq x/bfrac1n.$$




analysis logarithms




share|cite|improve this question

















  • 2




    Do you know/can you show that $$sum_n leqslant y frac1n = log y + gamma + Obiggl(frac1ybiggr),?$$
    – Daniel Fischer♦
    Jul 16 at 13:17












up vote
2
down vote

favorite









up vote
2
down vote

favorite











Trying to understand a paper which makes use of the following "fact" about logarithms. Can someone please explain why it is so?



Let $phi(x) = sum_x/a leq nleq x/b 1/n$. Then $lim_xto inftyphi(x) = log(a/b)$. In other words,
$$log(a/b) = lim_xto inftysum_x/a leq nleq x/bfrac1n.$$




analysis logarithms




share|cite|improve this question













Trying to understand a paper which makes use of the following "fact" about logarithms. Can someone please explain why it is so?



Let $phi(x) = sum_x/a leq nleq x/b 1/n$. Then $lim_xto inftyphi(x) = log(a/b)$. In other words,
$$log(a/b) = lim_xto inftysum_x/a leq nleq x/bfrac1n.$$





analysis logarithms





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share|cite|improve this question




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edited Jul 16 at 13:29









Parcly Taxel

33.6k136588




33.6k136588









asked Jul 16 at 13:10









abe.nong

214111




214111







  • 2




    Do you know/can you show that $$sum_n leqslant y frac1n = log y + gamma + Obiggl(frac1ybiggr),?$$
    – Daniel Fischer♦
    Jul 16 at 13:17












  • 2




    Do you know/can you show that $$sum_n leqslant y frac1n = log y + gamma + Obiggl(frac1ybiggr),?$$
    – Daniel Fischer♦
    Jul 16 at 13:17







2




2




Do you know/can you show that $$sum_n leqslant y frac1n = log y + gamma + Obiggl(frac1ybiggr),?$$
– Daniel Fischer♦
Jul 16 at 13:17




Do you know/can you show that $$sum_n leqslant y frac1n = log y + gamma + Obiggl(frac1ybiggr),?$$
– Daniel Fischer♦
Jul 16 at 13:17










1 Answer
1






active

oldest

votes

















up vote
2
down vote



accepted










It is known that
$$lim_ntoinftyleft(sum_i=1^nfrac1i-ln nright)=gamma$$
Therefore
$$lim_xtoinftysum_x/ale nle x/bfrac 1n=lim_xtoinftyleft(sum_i=1^lfloor x/brfloorfrac1i-sum_i=1^lceil x/arceilfrac1iright)=lim_xtoinfty(lnlfloor x/brfloor+gamma-lnlceil x/arceil-gamma)$$
$$=lim_xtoinftylnfraclfloor x/brfloorlceil x/arceil=lnfrac ab$$






share|cite|improve this answer





















  • ah in the sum after the first equality, do you mean to switch b and a? other than that, thanks heaps!
    – abe.nong
    Jul 16 at 13:38










  • @abe.nong it must have been a typo or slip-up of yours.
    – Parcly Taxel
    Jul 16 at 13:39










  • it definitely was my mistake. thanks!
    – abe.nong
    Jul 16 at 13:58










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1 Answer
1






active

oldest

votes








1 Answer
1






active

oldest

votes









active

oldest

votes






active

oldest

votes








up vote
2
down vote



accepted










It is known that
$$lim_ntoinftyleft(sum_i=1^nfrac1i-ln nright)=gamma$$
Therefore
$$lim_xtoinftysum_x/ale nle x/bfrac 1n=lim_xtoinftyleft(sum_i=1^lfloor x/brfloorfrac1i-sum_i=1^lceil x/arceilfrac1iright)=lim_xtoinfty(lnlfloor x/brfloor+gamma-lnlceil x/arceil-gamma)$$
$$=lim_xtoinftylnfraclfloor x/brfloorlceil x/arceil=lnfrac ab$$






share|cite|improve this answer





















  • ah in the sum after the first equality, do you mean to switch b and a? other than that, thanks heaps!
    – abe.nong
    Jul 16 at 13:38










  • @abe.nong it must have been a typo or slip-up of yours.
    – Parcly Taxel
    Jul 16 at 13:39










  • it definitely was my mistake. thanks!
    – abe.nong
    Jul 16 at 13:58














up vote
2
down vote



accepted










It is known that
$$lim_ntoinftyleft(sum_i=1^nfrac1i-ln nright)=gamma$$
Therefore
$$lim_xtoinftysum_x/ale nle x/bfrac 1n=lim_xtoinftyleft(sum_i=1^lfloor x/brfloorfrac1i-sum_i=1^lceil x/arceilfrac1iright)=lim_xtoinfty(lnlfloor x/brfloor+gamma-lnlceil x/arceil-gamma)$$
$$=lim_xtoinftylnfraclfloor x/brfloorlceil x/arceil=lnfrac ab$$






share|cite|improve this answer





















  • ah in the sum after the first equality, do you mean to switch b and a? other than that, thanks heaps!
    – abe.nong
    Jul 16 at 13:38










  • @abe.nong it must have been a typo or slip-up of yours.
    – Parcly Taxel
    Jul 16 at 13:39










  • it definitely was my mistake. thanks!
    – abe.nong
    Jul 16 at 13:58












up vote
2
down vote



accepted







up vote
2
down vote



accepted






It is known that
$$lim_ntoinftyleft(sum_i=1^nfrac1i-ln nright)=gamma$$
Therefore
$$lim_xtoinftysum_x/ale nle x/bfrac 1n=lim_xtoinftyleft(sum_i=1^lfloor x/brfloorfrac1i-sum_i=1^lceil x/arceilfrac1iright)=lim_xtoinfty(lnlfloor x/brfloor+gamma-lnlceil x/arceil-gamma)$$
$$=lim_xtoinftylnfraclfloor x/brfloorlceil x/arceil=lnfrac ab$$






share|cite|improve this answer













It is known that
$$lim_ntoinftyleft(sum_i=1^nfrac1i-ln nright)=gamma$$
Therefore
$$lim_xtoinftysum_x/ale nle x/bfrac 1n=lim_xtoinftyleft(sum_i=1^lfloor x/brfloorfrac1i-sum_i=1^lceil x/arceilfrac1iright)=lim_xtoinfty(lnlfloor x/brfloor+gamma-lnlceil x/arceil-gamma)$$
$$=lim_xtoinftylnfraclfloor x/brfloorlceil x/arceil=lnfrac ab$$







share|cite|improve this answer













share|cite|improve this answer



share|cite|improve this answer











answered Jul 16 at 13:26









Parcly Taxel

33.6k136588




33.6k136588











  • ah in the sum after the first equality, do you mean to switch b and a? other than that, thanks heaps!
    – abe.nong
    Jul 16 at 13:38










  • @abe.nong it must have been a typo or slip-up of yours.
    – Parcly Taxel
    Jul 16 at 13:39










  • it definitely was my mistake. thanks!
    – abe.nong
    Jul 16 at 13:58
















  • ah in the sum after the first equality, do you mean to switch b and a? other than that, thanks heaps!
    – abe.nong
    Jul 16 at 13:38










  • @abe.nong it must have been a typo or slip-up of yours.
    – Parcly Taxel
    Jul 16 at 13:39










  • it definitely was my mistake. thanks!
    – abe.nong
    Jul 16 at 13:58















ah in the sum after the first equality, do you mean to switch b and a? other than that, thanks heaps!
– abe.nong
Jul 16 at 13:38




ah in the sum after the first equality, do you mean to switch b and a? other than that, thanks heaps!
– abe.nong
Jul 16 at 13:38












@abe.nong it must have been a typo or slip-up of yours.
– Parcly Taxel
Jul 16 at 13:39




@abe.nong it must have been a typo or slip-up of yours.
– Parcly Taxel
Jul 16 at 13:39












it definitely was my mistake. thanks!
– abe.nong
Jul 16 at 13:58




it definitely was my mistake. thanks!
– abe.nong
Jul 16 at 13:58












 

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