Analytic continuation of harmonic series












1














Is there an accepted analytic continuation of $sum_{n=1}^m frac{1}{n}$? Even a continuation to positive reals would be of interested, though negative and complex arguments would also be interesting.



I don't have a specific application in mind, but I'd very much like to understand how / if such a continuation could be accomplished. I've Googled but haven't come up with anything meaningful - perhaps because it's not possible?






sequences-and-series harmonic-functions harmonic-numbers analytic-continuation







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  • 1




    As is the case with the Gamma function, it would help to specify what properties you want the analytic continuation to retain. For instance, Euler tells us that $H_n=int_0^1 frac {1-x^n}{1-x}dx$. Replacing $n$ by a continuous parameter gives you an analytic continuation (at least for $n>0$). Is it useful?
    – lulu
    1 hour ago


















1














Is there an accepted analytic continuation of $sum_{n=1}^m frac{1}{n}$? Even a continuation to positive reals would be of interested, though negative and complex arguments would also be interesting.



I don't have a specific application in mind, but I'd very much like to understand how / if such a continuation could be accomplished. I've Googled but haven't come up with anything meaningful - perhaps because it's not possible?






sequences-and-series harmonic-functions harmonic-numbers analytic-continuation







share|cite|improve this question


















  • 1




    As is the case with the Gamma function, it would help to specify what properties you want the analytic continuation to retain. For instance, Euler tells us that $H_n=int_0^1 frac {1-x^n}{1-x}dx$. Replacing $n$ by a continuous parameter gives you an analytic continuation (at least for $n>0$). Is it useful?
    – lulu
    1 hour ago
















1












1








1







Is there an accepted analytic continuation of $sum_{n=1}^m frac{1}{n}$? Even a continuation to positive reals would be of interested, though negative and complex arguments would also be interesting.



I don't have a specific application in mind, but I'd very much like to understand how / if such a continuation could be accomplished. I've Googled but haven't come up with anything meaningful - perhaps because it's not possible?






sequences-and-series harmonic-functions harmonic-numbers analytic-continuation







share|cite|improve this question













Is there an accepted analytic continuation of $sum_{n=1}^m frac{1}{n}$? Even a continuation to positive reals would be of interested, though negative and complex arguments would also be interesting.



I don't have a specific application in mind, but I'd very much like to understand how / if such a continuation could be accomplished. I've Googled but haven't come up with anything meaningful - perhaps because it's not possible?






sequences-and-series harmonic-functions harmonic-numbers analytic-continuation




sequences-and-series harmonic-functions harmonic-numbers analytic-continuation






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asked 1 hour ago









Richard Burke-Ward

3068




3068








  • 1




    As is the case with the Gamma function, it would help to specify what properties you want the analytic continuation to retain. For instance, Euler tells us that $H_n=int_0^1 frac {1-x^n}{1-x}dx$. Replacing $n$ by a continuous parameter gives you an analytic continuation (at least for $n>0$). Is it useful?
    – lulu
    1 hour ago
















  • 1




    As is the case with the Gamma function, it would help to specify what properties you want the analytic continuation to retain. For instance, Euler tells us that $H_n=int_0^1 frac {1-x^n}{1-x}dx$. Replacing $n$ by a continuous parameter gives you an analytic continuation (at least for $n>0$). Is it useful?
    – lulu
    1 hour ago










1




1




As is the case with the Gamma function, it would help to specify what properties you want the analytic continuation to retain. For instance, Euler tells us that $H_n=int_0^1 frac {1-x^n}{1-x}dx$. Replacing $n$ by a continuous parameter gives you an analytic continuation (at least for $n>0$). Is it useful?
– lulu
1 hour ago






As is the case with the Gamma function, it would help to specify what properties you want the analytic continuation to retain. For instance, Euler tells us that $H_n=int_0^1 frac {1-x^n}{1-x}dx$. Replacing $n$ by a continuous parameter gives you an analytic continuation (at least for $n>0$). Is it useful?
– lulu
1 hour ago












2 Answers
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I am not sure if this is what you meant, but Wolfram Alpha has an analytic formula for the $n^{text{th}}$ harmonic number:



enter image description here



Here, the digamma function is $psi_0(x)=frac{Gamma'(x)}{Gamma(x)}$, which I believe is defined for all numbers in the complex plane except for negative real integers.






share|cite|improve this answer





























    0














    Let's try it in an elementary manner




    1. You can use the defining recursion valid for $nin Z^{+}$


    $$H_{n} = H_{n-1} + frac{1}{n}, H_{1}=1tag{1a}$$



    also for any complex $z$



    $$H_{z} = H_{z-1} + frac{1}{z}, H_{1}=1tag{1b}$$



    For instance for $z=1$ we obtain $$H_{1} = H_{0} + frac{1}{1}$$



    from which we conclude that $H_{0}=0$.



    If we try to find $H_{-1}$ we encounter the problem that from $H_0 = 0 = lim_{zto0}(H_{-1+z} + frac{1}{z})$ we find that $H_{z} simeq frac{1}{z}$ for $zsimeq 0$. In other words, $H_{z}$ has a simple pole at $z=-1$.



    We could continue in this manner to go to further into the region of negative $z$ but we prefer the following approach.




    1. Starting with this formula for the harmonic number which is valid for $nin Z^{+}$


    $$H_{n} = frac{1}{2}+ ... + frac{1}{n}\\=frac{1}{1}+ frac{1}{2}+ ... + frac{1}{n} +frac{1}{1+n}+frac{1}{n+2} + ... \;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;-frac{1}{1+n}- frac{1}{n+2} + ...\=sum_{k=0}^infty left(frac{1}{k}-frac{1}{ (k+n)}right)tag{2}$$



    The sum can be written as



    $$H_{n}= sum_{k=0}^infty frac{n}{k (k+n)}tag{3}$$



    and this can be extended immediately to complex values $z$ in place of $n$



    $$H_{z}= sum_{k=0}^infty frac{z}{k (k+z)}=sum_{k=0}^infty left(frac{1}{k} -frac{1}{k+z}right)tag{4}$$



    This sum is convergent (the proof is left to the reader) for any $z$ except for $z=-1, -2, ...$ where $H_{z}$ has simple poles with residue $-1$.



    Hence $(4)$ gives the analytic continuation.



    For instance close to $z=0$ we have as in 1. that



    $$H_{z} simeq z sum_{k=0}^infty frac{1}{k^2} = z;zeta(2) =z;frac{pi^2}{6}to 0 $$



    We can also derive an integral representation from the second form of $(4)$ writing



    $$frac{1}{k} -frac{1}{k+z} =int_0^1 (x^{k-1}-x^{z+k-1}),dx $$



    Performing the sum under the integral is just doing a geometric sum and gives



    $$H_{z} = int_0^1 frac{1-x^{z}}{1-x},dx tag{5}$$






    share|cite|improve this answer





















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      2 Answers
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      2 Answers
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      3














      I am not sure if this is what you meant, but Wolfram Alpha has an analytic formula for the $n^{text{th}}$ harmonic number:



      enter image description here



      Here, the digamma function is $psi_0(x)=frac{Gamma'(x)}{Gamma(x)}$, which I believe is defined for all numbers in the complex plane except for negative real integers.






      share|cite|improve this answer


























        3














        I am not sure if this is what you meant, but Wolfram Alpha has an analytic formula for the $n^{text{th}}$ harmonic number:



        enter image description here



        Here, the digamma function is $psi_0(x)=frac{Gamma'(x)}{Gamma(x)}$, which I believe is defined for all numbers in the complex plane except for negative real integers.






        share|cite|improve this answer
























          3












          3








          3






          I am not sure if this is what you meant, but Wolfram Alpha has an analytic formula for the $n^{text{th}}$ harmonic number:



          enter image description here



          Here, the digamma function is $psi_0(x)=frac{Gamma'(x)}{Gamma(x)}$, which I believe is defined for all numbers in the complex plane except for negative real integers.






          share|cite|improve this answer












          I am not sure if this is what you meant, but Wolfram Alpha has an analytic formula for the $n^{text{th}}$ harmonic number:



          enter image description here



          Here, the digamma function is $psi_0(x)=frac{Gamma'(x)}{Gamma(x)}$, which I believe is defined for all numbers in the complex plane except for negative real integers.







          share|cite|improve this answer












          share|cite|improve this answer



          share|cite|improve this answer










          answered 1 hour ago









          Noble Mushtak

          14.7k1734




          14.7k1734























              0














              Let's try it in an elementary manner




              1. You can use the defining recursion valid for $nin Z^{+}$


              $$H_{n} = H_{n-1} + frac{1}{n}, H_{1}=1tag{1a}$$



              also for any complex $z$



              $$H_{z} = H_{z-1} + frac{1}{z}, H_{1}=1tag{1b}$$



              For instance for $z=1$ we obtain $$H_{1} = H_{0} + frac{1}{1}$$



              from which we conclude that $H_{0}=0$.



              If we try to find $H_{-1}$ we encounter the problem that from $H_0 = 0 = lim_{zto0}(H_{-1+z} + frac{1}{z})$ we find that $H_{z} simeq frac{1}{z}$ for $zsimeq 0$. In other words, $H_{z}$ has a simple pole at $z=-1$.



              We could continue in this manner to go to further into the region of negative $z$ but we prefer the following approach.




              1. Starting with this formula for the harmonic number which is valid for $nin Z^{+}$


              $$H_{n} = frac{1}{2}+ ... + frac{1}{n}\\=frac{1}{1}+ frac{1}{2}+ ... + frac{1}{n} +frac{1}{1+n}+frac{1}{n+2} + ... \;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;-frac{1}{1+n}- frac{1}{n+2} + ...\=sum_{k=0}^infty left(frac{1}{k}-frac{1}{ (k+n)}right)tag{2}$$



              The sum can be written as



              $$H_{n}= sum_{k=0}^infty frac{n}{k (k+n)}tag{3}$$



              and this can be extended immediately to complex values $z$ in place of $n$



              $$H_{z}= sum_{k=0}^infty frac{z}{k (k+z)}=sum_{k=0}^infty left(frac{1}{k} -frac{1}{k+z}right)tag{4}$$



              This sum is convergent (the proof is left to the reader) for any $z$ except for $z=-1, -2, ...$ where $H_{z}$ has simple poles with residue $-1$.



              Hence $(4)$ gives the analytic continuation.



              For instance close to $z=0$ we have as in 1. that



              $$H_{z} simeq z sum_{k=0}^infty frac{1}{k^2} = z;zeta(2) =z;frac{pi^2}{6}to 0 $$



              We can also derive an integral representation from the second form of $(4)$ writing



              $$frac{1}{k} -frac{1}{k+z} =int_0^1 (x^{k-1}-x^{z+k-1}),dx $$



              Performing the sum under the integral is just doing a geometric sum and gives



              $$H_{z} = int_0^1 frac{1-x^{z}}{1-x},dx tag{5}$$






              share|cite|improve this answer


























                0














                Let's try it in an elementary manner




                1. You can use the defining recursion valid for $nin Z^{+}$


                $$H_{n} = H_{n-1} + frac{1}{n}, H_{1}=1tag{1a}$$



                also for any complex $z$



                $$H_{z} = H_{z-1} + frac{1}{z}, H_{1}=1tag{1b}$$



                For instance for $z=1$ we obtain $$H_{1} = H_{0} + frac{1}{1}$$



                from which we conclude that $H_{0}=0$.



                If we try to find $H_{-1}$ we encounter the problem that from $H_0 = 0 = lim_{zto0}(H_{-1+z} + frac{1}{z})$ we find that $H_{z} simeq frac{1}{z}$ for $zsimeq 0$. In other words, $H_{z}$ has a simple pole at $z=-1$.



                We could continue in this manner to go to further into the region of negative $z$ but we prefer the following approach.




                1. Starting with this formula for the harmonic number which is valid for $nin Z^{+}$


                $$H_{n} = frac{1}{2}+ ... + frac{1}{n}\\=frac{1}{1}+ frac{1}{2}+ ... + frac{1}{n} +frac{1}{1+n}+frac{1}{n+2} + ... \;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;-frac{1}{1+n}- frac{1}{n+2} + ...\=sum_{k=0}^infty left(frac{1}{k}-frac{1}{ (k+n)}right)tag{2}$$



                The sum can be written as



                $$H_{n}= sum_{k=0}^infty frac{n}{k (k+n)}tag{3}$$



                and this can be extended immediately to complex values $z$ in place of $n$



                $$H_{z}= sum_{k=0}^infty frac{z}{k (k+z)}=sum_{k=0}^infty left(frac{1}{k} -frac{1}{k+z}right)tag{4}$$



                This sum is convergent (the proof is left to the reader) for any $z$ except for $z=-1, -2, ...$ where $H_{z}$ has simple poles with residue $-1$.



                Hence $(4)$ gives the analytic continuation.



                For instance close to $z=0$ we have as in 1. that



                $$H_{z} simeq z sum_{k=0}^infty frac{1}{k^2} = z;zeta(2) =z;frac{pi^2}{6}to 0 $$



                We can also derive an integral representation from the second form of $(4)$ writing



                $$frac{1}{k} -frac{1}{k+z} =int_0^1 (x^{k-1}-x^{z+k-1}),dx $$



                Performing the sum under the integral is just doing a geometric sum and gives



                $$H_{z} = int_0^1 frac{1-x^{z}}{1-x},dx tag{5}$$






                share|cite|improve this answer
























                  0












                  0








                  0






                  Let's try it in an elementary manner




                  1. You can use the defining recursion valid for $nin Z^{+}$


                  $$H_{n} = H_{n-1} + frac{1}{n}, H_{1}=1tag{1a}$$



                  also for any complex $z$



                  $$H_{z} = H_{z-1} + frac{1}{z}, H_{1}=1tag{1b}$$



                  For instance for $z=1$ we obtain $$H_{1} = H_{0} + frac{1}{1}$$



                  from which we conclude that $H_{0}=0$.



                  If we try to find $H_{-1}$ we encounter the problem that from $H_0 = 0 = lim_{zto0}(H_{-1+z} + frac{1}{z})$ we find that $H_{z} simeq frac{1}{z}$ for $zsimeq 0$. In other words, $H_{z}$ has a simple pole at $z=-1$.



                  We could continue in this manner to go to further into the region of negative $z$ but we prefer the following approach.




                  1. Starting with this formula for the harmonic number which is valid for $nin Z^{+}$


                  $$H_{n} = frac{1}{2}+ ... + frac{1}{n}\\=frac{1}{1}+ frac{1}{2}+ ... + frac{1}{n} +frac{1}{1+n}+frac{1}{n+2} + ... \;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;-frac{1}{1+n}- frac{1}{n+2} + ...\=sum_{k=0}^infty left(frac{1}{k}-frac{1}{ (k+n)}right)tag{2}$$



                  The sum can be written as



                  $$H_{n}= sum_{k=0}^infty frac{n}{k (k+n)}tag{3}$$



                  and this can be extended immediately to complex values $z$ in place of $n$



                  $$H_{z}= sum_{k=0}^infty frac{z}{k (k+z)}=sum_{k=0}^infty left(frac{1}{k} -frac{1}{k+z}right)tag{4}$$



                  This sum is convergent (the proof is left to the reader) for any $z$ except for $z=-1, -2, ...$ where $H_{z}$ has simple poles with residue $-1$.



                  Hence $(4)$ gives the analytic continuation.



                  For instance close to $z=0$ we have as in 1. that



                  $$H_{z} simeq z sum_{k=0}^infty frac{1}{k^2} = z;zeta(2) =z;frac{pi^2}{6}to 0 $$



                  We can also derive an integral representation from the second form of $(4)$ writing



                  $$frac{1}{k} -frac{1}{k+z} =int_0^1 (x^{k-1}-x^{z+k-1}),dx $$



                  Performing the sum under the integral is just doing a geometric sum and gives



                  $$H_{z} = int_0^1 frac{1-x^{z}}{1-x},dx tag{5}$$






                  share|cite|improve this answer












                  Let's try it in an elementary manner




                  1. You can use the defining recursion valid for $nin Z^{+}$


                  $$H_{n} = H_{n-1} + frac{1}{n}, H_{1}=1tag{1a}$$



                  also for any complex $z$



                  $$H_{z} = H_{z-1} + frac{1}{z}, H_{1}=1tag{1b}$$



                  For instance for $z=1$ we obtain $$H_{1} = H_{0} + frac{1}{1}$$



                  from which we conclude that $H_{0}=0$.



                  If we try to find $H_{-1}$ we encounter the problem that from $H_0 = 0 = lim_{zto0}(H_{-1+z} + frac{1}{z})$ we find that $H_{z} simeq frac{1}{z}$ for $zsimeq 0$. In other words, $H_{z}$ has a simple pole at $z=-1$.



                  We could continue in this manner to go to further into the region of negative $z$ but we prefer the following approach.




                  1. Starting with this formula for the harmonic number which is valid for $nin Z^{+}$


                  $$H_{n} = frac{1}{2}+ ... + frac{1}{n}\\=frac{1}{1}+ frac{1}{2}+ ... + frac{1}{n} +frac{1}{1+n}+frac{1}{n+2} + ... \;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;-frac{1}{1+n}- frac{1}{n+2} + ...\=sum_{k=0}^infty left(frac{1}{k}-frac{1}{ (k+n)}right)tag{2}$$



                  The sum can be written as



                  $$H_{n}= sum_{k=0}^infty frac{n}{k (k+n)}tag{3}$$



                  and this can be extended immediately to complex values $z$ in place of $n$



                  $$H_{z}= sum_{k=0}^infty frac{z}{k (k+z)}=sum_{k=0}^infty left(frac{1}{k} -frac{1}{k+z}right)tag{4}$$



                  This sum is convergent (the proof is left to the reader) for any $z$ except for $z=-1, -2, ...$ where $H_{z}$ has simple poles with residue $-1$.



                  Hence $(4)$ gives the analytic continuation.



                  For instance close to $z=0$ we have as in 1. that



                  $$H_{z} simeq z sum_{k=0}^infty frac{1}{k^2} = z;zeta(2) =z;frac{pi^2}{6}to 0 $$



                  We can also derive an integral representation from the second form of $(4)$ writing



                  $$frac{1}{k} -frac{1}{k+z} =int_0^1 (x^{k-1}-x^{z+k-1}),dx $$



                  Performing the sum under the integral is just doing a geometric sum and gives



                  $$H_{z} = int_0^1 frac{1-x^{z}}{1-x},dx tag{5}$$







                  share|cite|improve this answer












                  share|cite|improve this answer



                  share|cite|improve this answer










                  answered 24 mins ago









                  Dr. Wolfgang Hintze

                  3,150617




                  3,150617






























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