Sums of Reciprocals of Polynomials and Harmonic Numbers












2












$begingroup$


This is question based on a pattern I have noticed while using mathematica. Let $P(x)$ be a polynomial with real, simple, negative roots $r_n$ ($n:1,2,...,k$) and define



$$Q_n=lim_{xto r_n}frac{P(x)}{x-r_n}.$$



Then the pattern seems to be



$$sum_{n=1}^{infty}frac{1}{P(n)}=-sum_{i=1}^kfrac{H(-r_i)}{Q_i}$$



where $H(x)$ denotes the analytic continuation of the harmonic nubmers:



$$H(x)=int_0^1frac{1-t^x}{1-t}dt.$$



For example, for $P(x)=(x+frac{1}{2})(x+pi)(x+e)$, we have



$$sum_{n=1}^{infty}frac{1}{(x+frac{1}{2})(x+pi)(x+e)}=0.0841731$$



while



$$-frac{H(1/2)}{(1/2-pi)(1/2-e)}-frac{H(-pi)}{(pi-1/2)(pi-e)}-frac{H(-e)}{(e-1/2)(e-pi)}=0.0841731.$$



For clarification, the reason the roots must be negative is to sidestep any convergence issues in the infinite sum. My question is, does anyone know where I can find a proof of this fact or what some good topics/terms might be to search for? I have been handily using this fact in my current research, but upon starting the writeup today, I realized that I had never actually proved it.



If it helps, I believe this source deals with a more generalized version of this problem (as the digamma function and harmonic number are intimately linked). Unfortunately, they did not include a description of their coefficients (as far as I can tell) and only showed that it was a combination of a finite number of digamma functions.






sequences-and-series polynomials harmonic-numbers digamma-function







share|cite|improve this question









$endgroup$

















    2












    $begingroup$


    This is question based on a pattern I have noticed while using mathematica. Let $P(x)$ be a polynomial with real, simple, negative roots $r_n$ ($n:1,2,...,k$) and define



    $$Q_n=lim_{xto r_n}frac{P(x)}{x-r_n}.$$



    Then the pattern seems to be



    $$sum_{n=1}^{infty}frac{1}{P(n)}=-sum_{i=1}^kfrac{H(-r_i)}{Q_i}$$



    where $H(x)$ denotes the analytic continuation of the harmonic nubmers:



    $$H(x)=int_0^1frac{1-t^x}{1-t}dt.$$



    For example, for $P(x)=(x+frac{1}{2})(x+pi)(x+e)$, we have



    $$sum_{n=1}^{infty}frac{1}{(x+frac{1}{2})(x+pi)(x+e)}=0.0841731$$



    while



    $$-frac{H(1/2)}{(1/2-pi)(1/2-e)}-frac{H(-pi)}{(pi-1/2)(pi-e)}-frac{H(-e)}{(e-1/2)(e-pi)}=0.0841731.$$



    For clarification, the reason the roots must be negative is to sidestep any convergence issues in the infinite sum. My question is, does anyone know where I can find a proof of this fact or what some good topics/terms might be to search for? I have been handily using this fact in my current research, but upon starting the writeup today, I realized that I had never actually proved it.



    If it helps, I believe this source deals with a more generalized version of this problem (as the digamma function and harmonic number are intimately linked). Unfortunately, they did not include a description of their coefficients (as far as I can tell) and only showed that it was a combination of a finite number of digamma functions.






    sequences-and-series polynomials harmonic-numbers digamma-function







    share|cite|improve this question









    $endgroup$















      2












      2








      2





      $begingroup$


      This is question based on a pattern I have noticed while using mathematica. Let $P(x)$ be a polynomial with real, simple, negative roots $r_n$ ($n:1,2,...,k$) and define



      $$Q_n=lim_{xto r_n}frac{P(x)}{x-r_n}.$$



      Then the pattern seems to be



      $$sum_{n=1}^{infty}frac{1}{P(n)}=-sum_{i=1}^kfrac{H(-r_i)}{Q_i}$$



      where $H(x)$ denotes the analytic continuation of the harmonic nubmers:



      $$H(x)=int_0^1frac{1-t^x}{1-t}dt.$$



      For example, for $P(x)=(x+frac{1}{2})(x+pi)(x+e)$, we have



      $$sum_{n=1}^{infty}frac{1}{(x+frac{1}{2})(x+pi)(x+e)}=0.0841731$$



      while



      $$-frac{H(1/2)}{(1/2-pi)(1/2-e)}-frac{H(-pi)}{(pi-1/2)(pi-e)}-frac{H(-e)}{(e-1/2)(e-pi)}=0.0841731.$$



      For clarification, the reason the roots must be negative is to sidestep any convergence issues in the infinite sum. My question is, does anyone know where I can find a proof of this fact or what some good topics/terms might be to search for? I have been handily using this fact in my current research, but upon starting the writeup today, I realized that I had never actually proved it.



      If it helps, I believe this source deals with a more generalized version of this problem (as the digamma function and harmonic number are intimately linked). Unfortunately, they did not include a description of their coefficients (as far as I can tell) and only showed that it was a combination of a finite number of digamma functions.






      sequences-and-series polynomials harmonic-numbers digamma-function







      share|cite|improve this question









      $endgroup$




      This is question based on a pattern I have noticed while using mathematica. Let $P(x)$ be a polynomial with real, simple, negative roots $r_n$ ($n:1,2,...,k$) and define



      $$Q_n=lim_{xto r_n}frac{P(x)}{x-r_n}.$$



      Then the pattern seems to be



      $$sum_{n=1}^{infty}frac{1}{P(n)}=-sum_{i=1}^kfrac{H(-r_i)}{Q_i}$$



      where $H(x)$ denotes the analytic continuation of the harmonic nubmers:



      $$H(x)=int_0^1frac{1-t^x}{1-t}dt.$$



      For example, for $P(x)=(x+frac{1}{2})(x+pi)(x+e)$, we have



      $$sum_{n=1}^{infty}frac{1}{(x+frac{1}{2})(x+pi)(x+e)}=0.0841731$$



      while



      $$-frac{H(1/2)}{(1/2-pi)(1/2-e)}-frac{H(-pi)}{(pi-1/2)(pi-e)}-frac{H(-e)}{(e-1/2)(e-pi)}=0.0841731.$$



      For clarification, the reason the roots must be negative is to sidestep any convergence issues in the infinite sum. My question is, does anyone know where I can find a proof of this fact or what some good topics/terms might be to search for? I have been handily using this fact in my current research, but upon starting the writeup today, I realized that I had never actually proved it.



      If it helps, I believe this source deals with a more generalized version of this problem (as the digamma function and harmonic number are intimately linked). Unfortunately, they did not include a description of their coefficients (as far as I can tell) and only showed that it was a combination of a finite number of digamma functions.






      sequences-and-series polynomials harmonic-numbers digamma-function




      sequences-and-series polynomials harmonic-numbers digamma-function






      share|cite|improve this question













      share|cite|improve this question











      share|cite|improve this question




      share|cite|improve this question










      asked Jan 11 at 15:37









      Nick GuerreroNick Guerrero

      544411




      544411






















          1 Answer
          1






          active

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          3












          $begingroup$

          Using the reciprocals of your $Q_n$,
          $$
          begin{align}
          Q_n
          &=lim_{xto r_n}frac{x-r_n}{P(x)}
          end{align}
          $$
          The Heaviside Method for Partial Fractions says.
          $$
          frac1{P(x)}=sum_{n=1}^kfrac{Q_n}{x-r_n}
          $$
          Note that if $kge2$,
          $$
          begin{align}
          sum_{n=1}^kQ_n
          &=lim_{xtoinfty}sum_{n=1}^kfrac{x,Q_n}{x-r_n}\
          &=lim_{xtoinfty}frac{x}{P(x)}\[6pt]
          &=0
          end{align}
          $$
          Therefore,
          $$
          begin{align}
          sum_{j=1}^inftyfrac1{P(j)}
          &=sum_{j=1}^inftysum_{n=1}^kfrac{Q_n}{j-r_n}\
          &=sum_{j=1}^inftysum_{n=1}^kQ_nleft(frac1{j-r_n}-frac1jright)\
          &=sum_{n=1}^kQ_nsum_{j=1}^inftyleft(frac1{j-r_n}-frac1jright)\
          &=-sum_{n=1}^kQ_nH(-r_n)
          end{align}
          $$






          share|cite|improve this answer











          $endgroup$













          • $begingroup$
            Perfect, this is exactly what I was looking for
            $endgroup$
            – Nick Guerrero
            Jan 11 at 17:15











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          1 Answer
          1






          active

          oldest

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          1 Answer
          1






          active

          oldest

          votes









          active

          oldest

          votes






          active

          oldest

          votes









          3












          $begingroup$

          Using the reciprocals of your $Q_n$,
          $$
          begin{align}
          Q_n
          &=lim_{xto r_n}frac{x-r_n}{P(x)}
          end{align}
          $$
          The Heaviside Method for Partial Fractions says.
          $$
          frac1{P(x)}=sum_{n=1}^kfrac{Q_n}{x-r_n}
          $$
          Note that if $kge2$,
          $$
          begin{align}
          sum_{n=1}^kQ_n
          &=lim_{xtoinfty}sum_{n=1}^kfrac{x,Q_n}{x-r_n}\
          &=lim_{xtoinfty}frac{x}{P(x)}\[6pt]
          &=0
          end{align}
          $$
          Therefore,
          $$
          begin{align}
          sum_{j=1}^inftyfrac1{P(j)}
          &=sum_{j=1}^inftysum_{n=1}^kfrac{Q_n}{j-r_n}\
          &=sum_{j=1}^inftysum_{n=1}^kQ_nleft(frac1{j-r_n}-frac1jright)\
          &=sum_{n=1}^kQ_nsum_{j=1}^inftyleft(frac1{j-r_n}-frac1jright)\
          &=-sum_{n=1}^kQ_nH(-r_n)
          end{align}
          $$






          share|cite|improve this answer











          $endgroup$













          • $begingroup$
            Perfect, this is exactly what I was looking for
            $endgroup$
            – Nick Guerrero
            Jan 11 at 17:15
















          3












          $begingroup$

          Using the reciprocals of your $Q_n$,
          $$
          begin{align}
          Q_n
          &=lim_{xto r_n}frac{x-r_n}{P(x)}
          end{align}
          $$
          The Heaviside Method for Partial Fractions says.
          $$
          frac1{P(x)}=sum_{n=1}^kfrac{Q_n}{x-r_n}
          $$
          Note that if $kge2$,
          $$
          begin{align}
          sum_{n=1}^kQ_n
          &=lim_{xtoinfty}sum_{n=1}^kfrac{x,Q_n}{x-r_n}\
          &=lim_{xtoinfty}frac{x}{P(x)}\[6pt]
          &=0
          end{align}
          $$
          Therefore,
          $$
          begin{align}
          sum_{j=1}^inftyfrac1{P(j)}
          &=sum_{j=1}^inftysum_{n=1}^kfrac{Q_n}{j-r_n}\
          &=sum_{j=1}^inftysum_{n=1}^kQ_nleft(frac1{j-r_n}-frac1jright)\
          &=sum_{n=1}^kQ_nsum_{j=1}^inftyleft(frac1{j-r_n}-frac1jright)\
          &=-sum_{n=1}^kQ_nH(-r_n)
          end{align}
          $$






          share|cite|improve this answer











          $endgroup$













          • $begingroup$
            Perfect, this is exactly what I was looking for
            $endgroup$
            – Nick Guerrero
            Jan 11 at 17:15














          3












          3








          3





          $begingroup$

          Using the reciprocals of your $Q_n$,
          $$
          begin{align}
          Q_n
          &=lim_{xto r_n}frac{x-r_n}{P(x)}
          end{align}
          $$
          The Heaviside Method for Partial Fractions says.
          $$
          frac1{P(x)}=sum_{n=1}^kfrac{Q_n}{x-r_n}
          $$
          Note that if $kge2$,
          $$
          begin{align}
          sum_{n=1}^kQ_n
          &=lim_{xtoinfty}sum_{n=1}^kfrac{x,Q_n}{x-r_n}\
          &=lim_{xtoinfty}frac{x}{P(x)}\[6pt]
          &=0
          end{align}
          $$
          Therefore,
          $$
          begin{align}
          sum_{j=1}^inftyfrac1{P(j)}
          &=sum_{j=1}^inftysum_{n=1}^kfrac{Q_n}{j-r_n}\
          &=sum_{j=1}^inftysum_{n=1}^kQ_nleft(frac1{j-r_n}-frac1jright)\
          &=sum_{n=1}^kQ_nsum_{j=1}^inftyleft(frac1{j-r_n}-frac1jright)\
          &=-sum_{n=1}^kQ_nH(-r_n)
          end{align}
          $$






          share|cite|improve this answer











          $endgroup$



          Using the reciprocals of your $Q_n$,
          $$
          begin{align}
          Q_n
          &=lim_{xto r_n}frac{x-r_n}{P(x)}
          end{align}
          $$
          The Heaviside Method for Partial Fractions says.
          $$
          frac1{P(x)}=sum_{n=1}^kfrac{Q_n}{x-r_n}
          $$
          Note that if $kge2$,
          $$
          begin{align}
          sum_{n=1}^kQ_n
          &=lim_{xtoinfty}sum_{n=1}^kfrac{x,Q_n}{x-r_n}\
          &=lim_{xtoinfty}frac{x}{P(x)}\[6pt]
          &=0
          end{align}
          $$
          Therefore,
          $$
          begin{align}
          sum_{j=1}^inftyfrac1{P(j)}
          &=sum_{j=1}^inftysum_{n=1}^kfrac{Q_n}{j-r_n}\
          &=sum_{j=1}^inftysum_{n=1}^kQ_nleft(frac1{j-r_n}-frac1jright)\
          &=sum_{n=1}^kQ_nsum_{j=1}^inftyleft(frac1{j-r_n}-frac1jright)\
          &=-sum_{n=1}^kQ_nH(-r_n)
          end{align}
          $$







          share|cite|improve this answer














          share|cite|improve this answer



          share|cite|improve this answer








          edited Jan 12 at 20:35

























          answered Jan 11 at 16:13









          robjohn♦robjohn

          266k27305628




          266k27305628












          • $begingroup$
            Perfect, this is exactly what I was looking for
            $endgroup$
            – Nick Guerrero
            Jan 11 at 17:15


















          • $begingroup$
            Perfect, this is exactly what I was looking for
            $endgroup$
            – Nick Guerrero
            Jan 11 at 17:15
















          $begingroup$
          Perfect, this is exactly what I was looking for
          $endgroup$
          – Nick Guerrero
          Jan 11 at 17:15




          $begingroup$
          Perfect, this is exactly what I was looking for
          $endgroup$
          – Nick Guerrero
          Jan 11 at 17:15


















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