First order approximation of $zeta$(s) at s=1












3












$begingroup$


I was playing around with Wolfram Alpha.
I found one interesting thing when I asked it to evaluate this particular summation.$$Sigma_{n=1}^inftyfrac{1}{n^{1+10^{-
10}}}$$
It returned this$$ approx10^{10}+gamma$$
That's right the number itself plus the Euler–Mascheroni constant.
I tried it for various other values and it turned out to follow the same pattern. As the value got smaller the closer it followed that pattern.
Image of Wolfram Alpha result click here
So formally stating $$ lim _{s rightarrow0 } zeta(1+s) = 1/s+ gamma$$
Is this true?
And why is this true?
Could it be proven by taking the first order approximation of the zeta function.






sequences-and-series approximation riemann-zeta







share|cite|improve this question











$endgroup$












  • $begingroup$
    Better now after your edit ! I shall fix my answer. Cheers :-)
    $endgroup$
    – Claude Leibovici
    Jan 18 at 8:43










  • $begingroup$
    arxiv.org/pdf/1201.2633.pdf could be of interest.
    $endgroup$
    – Claude Leibovici
    Jan 18 at 8:53
















3












$begingroup$


I was playing around with Wolfram Alpha.
I found one interesting thing when I asked it to evaluate this particular summation.$$Sigma_{n=1}^inftyfrac{1}{n^{1+10^{-
10}}}$$
It returned this$$ approx10^{10}+gamma$$
That's right the number itself plus the Euler–Mascheroni constant.
I tried it for various other values and it turned out to follow the same pattern. As the value got smaller the closer it followed that pattern.
Image of Wolfram Alpha result click here
So formally stating $$ lim _{s rightarrow0 } zeta(1+s) = 1/s+ gamma$$
Is this true?
And why is this true?
Could it be proven by taking the first order approximation of the zeta function.






sequences-and-series approximation riemann-zeta







share|cite|improve this question











$endgroup$












  • $begingroup$
    Better now after your edit ! I shall fix my answer. Cheers :-)
    $endgroup$
    – Claude Leibovici
    Jan 18 at 8:43










  • $begingroup$
    arxiv.org/pdf/1201.2633.pdf could be of interest.
    $endgroup$
    – Claude Leibovici
    Jan 18 at 8:53














3












3








3





$begingroup$


I was playing around with Wolfram Alpha.
I found one interesting thing when I asked it to evaluate this particular summation.$$Sigma_{n=1}^inftyfrac{1}{n^{1+10^{-
10}}}$$
It returned this$$ approx10^{10}+gamma$$
That's right the number itself plus the Euler–Mascheroni constant.
I tried it for various other values and it turned out to follow the same pattern. As the value got smaller the closer it followed that pattern.
Image of Wolfram Alpha result click here
So formally stating $$ lim _{s rightarrow0 } zeta(1+s) = 1/s+ gamma$$
Is this true?
And why is this true?
Could it be proven by taking the first order approximation of the zeta function.






sequences-and-series approximation riemann-zeta







share|cite|improve this question











$endgroup$




I was playing around with Wolfram Alpha.
I found one interesting thing when I asked it to evaluate this particular summation.$$Sigma_{n=1}^inftyfrac{1}{n^{1+10^{-
10}}}$$
It returned this$$ approx10^{10}+gamma$$
That's right the number itself plus the Euler–Mascheroni constant.
I tried it for various other values and it turned out to follow the same pattern. As the value got smaller the closer it followed that pattern.
Image of Wolfram Alpha result click here
So formally stating $$ lim _{s rightarrow0 } zeta(1+s) = 1/s+ gamma$$
Is this true?
And why is this true?
Could it be proven by taking the first order approximation of the zeta function.






sequences-and-series approximation riemann-zeta




sequences-and-series approximation riemann-zeta






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Jan 18 at 8:42







Karthik

















asked Jan 18 at 8:29









KarthikKarthik

13212




13212












  • $begingroup$
    Better now after your edit ! I shall fix my answer. Cheers :-)
    $endgroup$
    – Claude Leibovici
    Jan 18 at 8:43










  • $begingroup$
    arxiv.org/pdf/1201.2633.pdf could be of interest.
    $endgroup$
    – Claude Leibovici
    Jan 18 at 8:53


















  • $begingroup$
    Better now after your edit ! I shall fix my answer. Cheers :-)
    $endgroup$
    – Claude Leibovici
    Jan 18 at 8:43










  • $begingroup$
    arxiv.org/pdf/1201.2633.pdf could be of interest.
    $endgroup$
    – Claude Leibovici
    Jan 18 at 8:53
















$begingroup$
Better now after your edit ! I shall fix my answer. Cheers :-)
$endgroup$
– Claude Leibovici
Jan 18 at 8:43




$begingroup$
Better now after your edit ! I shall fix my answer. Cheers :-)
$endgroup$
– Claude Leibovici
Jan 18 at 8:43












$begingroup$
arxiv.org/pdf/1201.2633.pdf could be of interest.
$endgroup$
– Claude Leibovici
Jan 18 at 8:53




$begingroup$
arxiv.org/pdf/1201.2633.pdf could be of interest.
$endgroup$
– Claude Leibovici
Jan 18 at 8:53










1 Answer
1






active

oldest

votes


















0












$begingroup$

$$sum_{n=1}^inftyfrac{1}{n^{1+epsilon}}=zeta (1+epsilon)=frac{1}{epsilon }+gamma -gamma _1 epsilon +frac{gamma _2 }{2}epsilon
^2+Oleft(epsilon ^3right)$$ where appear the generalized Stieltjes constants.






share|cite|improve this answer











$endgroup$













  • $begingroup$
    Refer to the comment by Claude for more details.
    $endgroup$
    – Karthik
    Jan 18 at 9:08






  • 1




    $begingroup$
    @JackD'Aurizio. Please, don't call me Sir. I shall work on this tomorrow morning. Cheers & thank for the problem.
    $endgroup$
    – Claude Leibovici
    Jan 18 at 17:19










  • $begingroup$
    $zeta(s) = sum_{n=1}^infty frac1n n^{1-s} = (s-1) int_1^infty ( sum_{n le x} frac1n) x^{-s}dx$ may help to show $lim_{s to 1} zeta(s) - frac1{s-1} $ $=lim_{s to 1} (s-1) int_1^infty ( sum_{n le x} frac1n- log(x)) x^{-s}dx=lim_{s to 1} (s-1) int_1^infty (gamma+o(1)) x^{-s}dx = lim_{s to 1} gamma+o(1) = gamma$
    $endgroup$
    – reuns
    Jan 18 at 19:14













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1 Answer
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oldest

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1 Answer
1






active

oldest

votes









active

oldest

votes






active

oldest

votes









0












$begingroup$

$$sum_{n=1}^inftyfrac{1}{n^{1+epsilon}}=zeta (1+epsilon)=frac{1}{epsilon }+gamma -gamma _1 epsilon +frac{gamma _2 }{2}epsilon
^2+Oleft(epsilon ^3right)$$ where appear the generalized Stieltjes constants.






share|cite|improve this answer











$endgroup$













  • $begingroup$
    Refer to the comment by Claude for more details.
    $endgroup$
    – Karthik
    Jan 18 at 9:08






  • 1




    $begingroup$
    @JackD'Aurizio. Please, don't call me Sir. I shall work on this tomorrow morning. Cheers & thank for the problem.
    $endgroup$
    – Claude Leibovici
    Jan 18 at 17:19










  • $begingroup$
    $zeta(s) = sum_{n=1}^infty frac1n n^{1-s} = (s-1) int_1^infty ( sum_{n le x} frac1n) x^{-s}dx$ may help to show $lim_{s to 1} zeta(s) - frac1{s-1} $ $=lim_{s to 1} (s-1) int_1^infty ( sum_{n le x} frac1n- log(x)) x^{-s}dx=lim_{s to 1} (s-1) int_1^infty (gamma+o(1)) x^{-s}dx = lim_{s to 1} gamma+o(1) = gamma$
    $endgroup$
    – reuns
    Jan 18 at 19:14


















0












$begingroup$

$$sum_{n=1}^inftyfrac{1}{n^{1+epsilon}}=zeta (1+epsilon)=frac{1}{epsilon }+gamma -gamma _1 epsilon +frac{gamma _2 }{2}epsilon
^2+Oleft(epsilon ^3right)$$ where appear the generalized Stieltjes constants.






share|cite|improve this answer











$endgroup$













  • $begingroup$
    Refer to the comment by Claude for more details.
    $endgroup$
    – Karthik
    Jan 18 at 9:08






  • 1




    $begingroup$
    @JackD'Aurizio. Please, don't call me Sir. I shall work on this tomorrow morning. Cheers & thank for the problem.
    $endgroup$
    – Claude Leibovici
    Jan 18 at 17:19










  • $begingroup$
    $zeta(s) = sum_{n=1}^infty frac1n n^{1-s} = (s-1) int_1^infty ( sum_{n le x} frac1n) x^{-s}dx$ may help to show $lim_{s to 1} zeta(s) - frac1{s-1} $ $=lim_{s to 1} (s-1) int_1^infty ( sum_{n le x} frac1n- log(x)) x^{-s}dx=lim_{s to 1} (s-1) int_1^infty (gamma+o(1)) x^{-s}dx = lim_{s to 1} gamma+o(1) = gamma$
    $endgroup$
    – reuns
    Jan 18 at 19:14
















0












0








0





$begingroup$

$$sum_{n=1}^inftyfrac{1}{n^{1+epsilon}}=zeta (1+epsilon)=frac{1}{epsilon }+gamma -gamma _1 epsilon +frac{gamma _2 }{2}epsilon
^2+Oleft(epsilon ^3right)$$ where appear the generalized Stieltjes constants.






share|cite|improve this answer











$endgroup$



$$sum_{n=1}^inftyfrac{1}{n^{1+epsilon}}=zeta (1+epsilon)=frac{1}{epsilon }+gamma -gamma _1 epsilon +frac{gamma _2 }{2}epsilon
^2+Oleft(epsilon ^3right)$$ where appear the generalized Stieltjes constants.







share|cite|improve this answer














share|cite|improve this answer



share|cite|improve this answer








edited Jan 18 at 8:44

























answered Jan 18 at 8:39









Claude LeiboviciClaude Leibovici

123k1157134




123k1157134












  • $begingroup$
    Refer to the comment by Claude for more details.
    $endgroup$
    – Karthik
    Jan 18 at 9:08






  • 1




    $begingroup$
    @JackD'Aurizio. Please, don't call me Sir. I shall work on this tomorrow morning. Cheers & thank for the problem.
    $endgroup$
    – Claude Leibovici
    Jan 18 at 17:19










  • $begingroup$
    $zeta(s) = sum_{n=1}^infty frac1n n^{1-s} = (s-1) int_1^infty ( sum_{n le x} frac1n) x^{-s}dx$ may help to show $lim_{s to 1} zeta(s) - frac1{s-1} $ $=lim_{s to 1} (s-1) int_1^infty ( sum_{n le x} frac1n- log(x)) x^{-s}dx=lim_{s to 1} (s-1) int_1^infty (gamma+o(1)) x^{-s}dx = lim_{s to 1} gamma+o(1) = gamma$
    $endgroup$
    – reuns
    Jan 18 at 19:14




















  • $begingroup$
    Refer to the comment by Claude for more details.
    $endgroup$
    – Karthik
    Jan 18 at 9:08






  • 1




    $begingroup$
    @JackD'Aurizio. Please, don't call me Sir. I shall work on this tomorrow morning. Cheers & thank for the problem.
    $endgroup$
    – Claude Leibovici
    Jan 18 at 17:19










  • $begingroup$
    $zeta(s) = sum_{n=1}^infty frac1n n^{1-s} = (s-1) int_1^infty ( sum_{n le x} frac1n) x^{-s}dx$ may help to show $lim_{s to 1} zeta(s) - frac1{s-1} $ $=lim_{s to 1} (s-1) int_1^infty ( sum_{n le x} frac1n- log(x)) x^{-s}dx=lim_{s to 1} (s-1) int_1^infty (gamma+o(1)) x^{-s}dx = lim_{s to 1} gamma+o(1) = gamma$
    $endgroup$
    – reuns
    Jan 18 at 19:14


















$begingroup$
Refer to the comment by Claude for more details.
$endgroup$
– Karthik
Jan 18 at 9:08




$begingroup$
Refer to the comment by Claude for more details.
$endgroup$
– Karthik
Jan 18 at 9:08




1




1




$begingroup$
@JackD'Aurizio. Please, don't call me Sir. I shall work on this tomorrow morning. Cheers & thank for the problem.
$endgroup$
– Claude Leibovici
Jan 18 at 17:19




$begingroup$
@JackD'Aurizio. Please, don't call me Sir. I shall work on this tomorrow morning. Cheers & thank for the problem.
$endgroup$
– Claude Leibovici
Jan 18 at 17:19












$begingroup$
$zeta(s) = sum_{n=1}^infty frac1n n^{1-s} = (s-1) int_1^infty ( sum_{n le x} frac1n) x^{-s}dx$ may help to show $lim_{s to 1} zeta(s) - frac1{s-1} $ $=lim_{s to 1} (s-1) int_1^infty ( sum_{n le x} frac1n- log(x)) x^{-s}dx=lim_{s to 1} (s-1) int_1^infty (gamma+o(1)) x^{-s}dx = lim_{s to 1} gamma+o(1) = gamma$
$endgroup$
– reuns
Jan 18 at 19:14






$begingroup$
$zeta(s) = sum_{n=1}^infty frac1n n^{1-s} = (s-1) int_1^infty ( sum_{n le x} frac1n) x^{-s}dx$ may help to show $lim_{s to 1} zeta(s) - frac1{s-1} $ $=lim_{s to 1} (s-1) int_1^infty ( sum_{n le x} frac1n- log(x)) x^{-s}dx=lim_{s to 1} (s-1) int_1^infty (gamma+o(1)) x^{-s}dx = lim_{s to 1} gamma+o(1) = gamma$
$endgroup$
– reuns
Jan 18 at 19:14




















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