Generalizing $sum_{n=1}^{infty}frac{H_n{2n choose n}}{2^{2n}(2n-1)}=2$












6












$begingroup$


I was looking at this paper on section [17],




$$sum_{n=1}^{infty}frac{H_n{2n choose n}}{2^{2n}(2n-1)}=2tag1$$




Let generalize $(1)$



$$sum_{n=1}^{infty}frac{H_n{2n choose n}}{2^{2n}(2n-1)(2n-3)(2n-5)cdots [2n-(2k+1)]}tag2$$



Where $kge 0$



I conjectured the closed form for $(2)$ to be




$$sum_{n=1}^{infty}frac{H_n{2n choose n}}{2^{2n}(2n-1)(2n-3)cdots [2n-(2k+1)]}=frac{2(-1)^k}{(2k+1)!!(2k+1)}tag3$$




Here are a first few values of $k=1,2$ and $3$



$$begin{align}
sum_{n=1}^{infty}frac{H_n{2n choose n}}{2^{2n}(2n-1)(2n-3)}&=-frac{2}{9}tag4\
sum_{n=1}^{infty}frac{H_n{2n choose n}}{2^{2n}(2n-1)(2n-3)(2n-5)}&=frac{2}{75}tag5\
sum_{n=1}^{infty}frac{H_n{2n choose n}}{2^{2n}(2n-1)(2n-3)(2n-5)(2n-7)}&=-frac{2}{735}tag6
end{align}$$



How do we go about to prove this conjecture $(3)?$










share|cite|improve this question











$endgroup$












  • $begingroup$
    How did you determine (4), (5) and (6)? Have you tested them numerically?
    $endgroup$
    – Richard
    Jan 2 at 18:21










  • $begingroup$
    I tested them numerically. it seems correct
    $endgroup$
    – user583851
    Jan 2 at 18:30
















6












$begingroup$


I was looking at this paper on section [17],




$$sum_{n=1}^{infty}frac{H_n{2n choose n}}{2^{2n}(2n-1)}=2tag1$$




Let generalize $(1)$



$$sum_{n=1}^{infty}frac{H_n{2n choose n}}{2^{2n}(2n-1)(2n-3)(2n-5)cdots [2n-(2k+1)]}tag2$$



Where $kge 0$



I conjectured the closed form for $(2)$ to be




$$sum_{n=1}^{infty}frac{H_n{2n choose n}}{2^{2n}(2n-1)(2n-3)cdots [2n-(2k+1)]}=frac{2(-1)^k}{(2k+1)!!(2k+1)}tag3$$




Here are a first few values of $k=1,2$ and $3$



$$begin{align}
sum_{n=1}^{infty}frac{H_n{2n choose n}}{2^{2n}(2n-1)(2n-3)}&=-frac{2}{9}tag4\
sum_{n=1}^{infty}frac{H_n{2n choose n}}{2^{2n}(2n-1)(2n-3)(2n-5)}&=frac{2}{75}tag5\
sum_{n=1}^{infty}frac{H_n{2n choose n}}{2^{2n}(2n-1)(2n-3)(2n-5)(2n-7)}&=-frac{2}{735}tag6
end{align}$$



How do we go about to prove this conjecture $(3)?$










share|cite|improve this question











$endgroup$












  • $begingroup$
    How did you determine (4), (5) and (6)? Have you tested them numerically?
    $endgroup$
    – Richard
    Jan 2 at 18:21










  • $begingroup$
    I tested them numerically. it seems correct
    $endgroup$
    – user583851
    Jan 2 at 18:30














6












6








6


4



$begingroup$


I was looking at this paper on section [17],




$$sum_{n=1}^{infty}frac{H_n{2n choose n}}{2^{2n}(2n-1)}=2tag1$$




Let generalize $(1)$



$$sum_{n=1}^{infty}frac{H_n{2n choose n}}{2^{2n}(2n-1)(2n-3)(2n-5)cdots [2n-(2k+1)]}tag2$$



Where $kge 0$



I conjectured the closed form for $(2)$ to be




$$sum_{n=1}^{infty}frac{H_n{2n choose n}}{2^{2n}(2n-1)(2n-3)cdots [2n-(2k+1)]}=frac{2(-1)^k}{(2k+1)!!(2k+1)}tag3$$




Here are a first few values of $k=1,2$ and $3$



$$begin{align}
sum_{n=1}^{infty}frac{H_n{2n choose n}}{2^{2n}(2n-1)(2n-3)}&=-frac{2}{9}tag4\
sum_{n=1}^{infty}frac{H_n{2n choose n}}{2^{2n}(2n-1)(2n-3)(2n-5)}&=frac{2}{75}tag5\
sum_{n=1}^{infty}frac{H_n{2n choose n}}{2^{2n}(2n-1)(2n-3)(2n-5)(2n-7)}&=-frac{2}{735}tag6
end{align}$$



How do we go about to prove this conjecture $(3)?$










share|cite|improve this question











$endgroup$




I was looking at this paper on section [17],




$$sum_{n=1}^{infty}frac{H_n{2n choose n}}{2^{2n}(2n-1)}=2tag1$$




Let generalize $(1)$



$$sum_{n=1}^{infty}frac{H_n{2n choose n}}{2^{2n}(2n-1)(2n-3)(2n-5)cdots [2n-(2k+1)]}tag2$$



Where $kge 0$



I conjectured the closed form for $(2)$ to be




$$sum_{n=1}^{infty}frac{H_n{2n choose n}}{2^{2n}(2n-1)(2n-3)cdots [2n-(2k+1)]}=frac{2(-1)^k}{(2k+1)!!(2k+1)}tag3$$




Here are a first few values of $k=1,2$ and $3$



$$begin{align}
sum_{n=1}^{infty}frac{H_n{2n choose n}}{2^{2n}(2n-1)(2n-3)}&=-frac{2}{9}tag4\
sum_{n=1}^{infty}frac{H_n{2n choose n}}{2^{2n}(2n-1)(2n-3)(2n-5)}&=frac{2}{75}tag5\
sum_{n=1}^{infty}frac{H_n{2n choose n}}{2^{2n}(2n-1)(2n-3)(2n-5)(2n-7)}&=-frac{2}{735}tag6
end{align}$$



How do we go about to prove this conjecture $(3)?$







sequences-and-series closed-form harmonic-numbers






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share|cite|improve this question













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share|cite|improve this question








edited Jan 2 at 18:03









mrtaurho

4,12621234




4,12621234










asked Jan 2 at 17:43









user583851user583851

49119




49119












  • $begingroup$
    How did you determine (4), (5) and (6)? Have you tested them numerically?
    $endgroup$
    – Richard
    Jan 2 at 18:21










  • $begingroup$
    I tested them numerically. it seems correct
    $endgroup$
    – user583851
    Jan 2 at 18:30


















  • $begingroup$
    How did you determine (4), (5) and (6)? Have you tested them numerically?
    $endgroup$
    – Richard
    Jan 2 at 18:21










  • $begingroup$
    I tested them numerically. it seems correct
    $endgroup$
    – user583851
    Jan 2 at 18:30
















$begingroup$
How did you determine (4), (5) and (6)? Have you tested them numerically?
$endgroup$
– Richard
Jan 2 at 18:21




$begingroup$
How did you determine (4), (5) and (6)? Have you tested them numerically?
$endgroup$
– Richard
Jan 2 at 18:21












$begingroup$
I tested them numerically. it seems correct
$endgroup$
– user583851
Jan 2 at 18:30




$begingroup$
I tested them numerically. it seems correct
$endgroup$
– user583851
Jan 2 at 18:30










1 Answer
1






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oldest

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5












$begingroup$

What about reindexing and induction? The terms $frac{1}{(2n-1)cdots(2n-2k-1)}$ have a nice telescopic structure: by the residue theorem
$$ frac{1}{(2n-1)(2n-3)cdots(2n-2k-1)}=(-1)^ksum_{h=0}^{k}frac{(-1)^h}{(2n-2h-1)}cdotfrac{1}{2^{k+1}(2h)!!(2k-2h)!!} $$
equals
$$ frac{(-1)^k}{2^{2k+1}k!} sum_{h=0}^{k}frac{(-1)^h}{(2n-2h-1)}binom{k}{h}.$$
The natural temptation is now to compute
$$ sum_{ngeq 1}frac{H_n}{4^{n}}binom{2n}{n}frac{1}{2n-2h-1}$$
through $frac{-log(1-z)}{1-z}=sum_{ngeq 1}H_n z^n$ and $frac{1}{4^n}binom{2n}{n}=frac{2}{pi}int_{0}^{pi/2}left(costhetaright)^{2n},dtheta$, multiply both sides by $(-1)^k binom{k}{h}$, sum over $h=0,1,ldots,k$ and finish by invoking Fubini's theorem (allowing to switch the integrals with respect to $dtheta$ and $dz$) and the Fourier series $sum_{mgeq 1}frac{cos(mvarphi)}{m}$ and $sum_{mgeq 1}frac{sin(mvarphi)}{m}$.



The only obstruction is that $frac{1}{2n-2h-1}=int_{0}^{1}z^nleft[frac{1}{2z^{h+3/2}}right],dz$ does not hold unconditionally: we would have been happier in having rising Pochhammer symbols rather than falling ones. On the other hand, reindexing fixes this issue. Since $binom{2n+2}{n+1} = frac{2(2n+1)}{n+1}binom{2n}{n}$, the original series can be written as



$$ sum_{ngeq 1}frac{H_n binom{2n}{n}}{4^n(2n-1)} = sum_{ngeq 0}frac{2H_{n+1}binom{2n}{n}}{4^{n+1}(n+1)}=-frac{1}{pi}int_{0}^{1}sum_{ngeq 0}int_{0}^{pi/2}z^nleft(costhetaright)^{2n}log(1-z),dtheta,dz $$
or
$$ -frac{1}{pi}int_{0}^{1}int_{0}^{pi/2}frac{log(1-z)}{1-zcos^2theta},dtheta,dz =-frac{1}{2}int_{0}^{1}frac{log(1-z)}{sqrt{1-z}},dz,$$
clearly given by a derivative of the Beta function. This approach works also by replacing $(2n-1)$ with $(2n-1)cdots(2n-2k-1)$, you just have to be careful in managing the involved constants depending on $k$.



On second thought, I have just applied the binomial transform in disguise.






share|cite|improve this answer











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    5












    $begingroup$

    What about reindexing and induction? The terms $frac{1}{(2n-1)cdots(2n-2k-1)}$ have a nice telescopic structure: by the residue theorem
    $$ frac{1}{(2n-1)(2n-3)cdots(2n-2k-1)}=(-1)^ksum_{h=0}^{k}frac{(-1)^h}{(2n-2h-1)}cdotfrac{1}{2^{k+1}(2h)!!(2k-2h)!!} $$
    equals
    $$ frac{(-1)^k}{2^{2k+1}k!} sum_{h=0}^{k}frac{(-1)^h}{(2n-2h-1)}binom{k}{h}.$$
    The natural temptation is now to compute
    $$ sum_{ngeq 1}frac{H_n}{4^{n}}binom{2n}{n}frac{1}{2n-2h-1}$$
    through $frac{-log(1-z)}{1-z}=sum_{ngeq 1}H_n z^n$ and $frac{1}{4^n}binom{2n}{n}=frac{2}{pi}int_{0}^{pi/2}left(costhetaright)^{2n},dtheta$, multiply both sides by $(-1)^k binom{k}{h}$, sum over $h=0,1,ldots,k$ and finish by invoking Fubini's theorem (allowing to switch the integrals with respect to $dtheta$ and $dz$) and the Fourier series $sum_{mgeq 1}frac{cos(mvarphi)}{m}$ and $sum_{mgeq 1}frac{sin(mvarphi)}{m}$.



    The only obstruction is that $frac{1}{2n-2h-1}=int_{0}^{1}z^nleft[frac{1}{2z^{h+3/2}}right],dz$ does not hold unconditionally: we would have been happier in having rising Pochhammer symbols rather than falling ones. On the other hand, reindexing fixes this issue. Since $binom{2n+2}{n+1} = frac{2(2n+1)}{n+1}binom{2n}{n}$, the original series can be written as



    $$ sum_{ngeq 1}frac{H_n binom{2n}{n}}{4^n(2n-1)} = sum_{ngeq 0}frac{2H_{n+1}binom{2n}{n}}{4^{n+1}(n+1)}=-frac{1}{pi}int_{0}^{1}sum_{ngeq 0}int_{0}^{pi/2}z^nleft(costhetaright)^{2n}log(1-z),dtheta,dz $$
    or
    $$ -frac{1}{pi}int_{0}^{1}int_{0}^{pi/2}frac{log(1-z)}{1-zcos^2theta},dtheta,dz =-frac{1}{2}int_{0}^{1}frac{log(1-z)}{sqrt{1-z}},dz,$$
    clearly given by a derivative of the Beta function. This approach works also by replacing $(2n-1)$ with $(2n-1)cdots(2n-2k-1)$, you just have to be careful in managing the involved constants depending on $k$.



    On second thought, I have just applied the binomial transform in disguise.






    share|cite|improve this answer











    $endgroup$


















      5












      $begingroup$

      What about reindexing and induction? The terms $frac{1}{(2n-1)cdots(2n-2k-1)}$ have a nice telescopic structure: by the residue theorem
      $$ frac{1}{(2n-1)(2n-3)cdots(2n-2k-1)}=(-1)^ksum_{h=0}^{k}frac{(-1)^h}{(2n-2h-1)}cdotfrac{1}{2^{k+1}(2h)!!(2k-2h)!!} $$
      equals
      $$ frac{(-1)^k}{2^{2k+1}k!} sum_{h=0}^{k}frac{(-1)^h}{(2n-2h-1)}binom{k}{h}.$$
      The natural temptation is now to compute
      $$ sum_{ngeq 1}frac{H_n}{4^{n}}binom{2n}{n}frac{1}{2n-2h-1}$$
      through $frac{-log(1-z)}{1-z}=sum_{ngeq 1}H_n z^n$ and $frac{1}{4^n}binom{2n}{n}=frac{2}{pi}int_{0}^{pi/2}left(costhetaright)^{2n},dtheta$, multiply both sides by $(-1)^k binom{k}{h}$, sum over $h=0,1,ldots,k$ and finish by invoking Fubini's theorem (allowing to switch the integrals with respect to $dtheta$ and $dz$) and the Fourier series $sum_{mgeq 1}frac{cos(mvarphi)}{m}$ and $sum_{mgeq 1}frac{sin(mvarphi)}{m}$.



      The only obstruction is that $frac{1}{2n-2h-1}=int_{0}^{1}z^nleft[frac{1}{2z^{h+3/2}}right],dz$ does not hold unconditionally: we would have been happier in having rising Pochhammer symbols rather than falling ones. On the other hand, reindexing fixes this issue. Since $binom{2n+2}{n+1} = frac{2(2n+1)}{n+1}binom{2n}{n}$, the original series can be written as



      $$ sum_{ngeq 1}frac{H_n binom{2n}{n}}{4^n(2n-1)} = sum_{ngeq 0}frac{2H_{n+1}binom{2n}{n}}{4^{n+1}(n+1)}=-frac{1}{pi}int_{0}^{1}sum_{ngeq 0}int_{0}^{pi/2}z^nleft(costhetaright)^{2n}log(1-z),dtheta,dz $$
      or
      $$ -frac{1}{pi}int_{0}^{1}int_{0}^{pi/2}frac{log(1-z)}{1-zcos^2theta},dtheta,dz =-frac{1}{2}int_{0}^{1}frac{log(1-z)}{sqrt{1-z}},dz,$$
      clearly given by a derivative of the Beta function. This approach works also by replacing $(2n-1)$ with $(2n-1)cdots(2n-2k-1)$, you just have to be careful in managing the involved constants depending on $k$.



      On second thought, I have just applied the binomial transform in disguise.






      share|cite|improve this answer











      $endgroup$
















        5












        5








        5





        $begingroup$

        What about reindexing and induction? The terms $frac{1}{(2n-1)cdots(2n-2k-1)}$ have a nice telescopic structure: by the residue theorem
        $$ frac{1}{(2n-1)(2n-3)cdots(2n-2k-1)}=(-1)^ksum_{h=0}^{k}frac{(-1)^h}{(2n-2h-1)}cdotfrac{1}{2^{k+1}(2h)!!(2k-2h)!!} $$
        equals
        $$ frac{(-1)^k}{2^{2k+1}k!} sum_{h=0}^{k}frac{(-1)^h}{(2n-2h-1)}binom{k}{h}.$$
        The natural temptation is now to compute
        $$ sum_{ngeq 1}frac{H_n}{4^{n}}binom{2n}{n}frac{1}{2n-2h-1}$$
        through $frac{-log(1-z)}{1-z}=sum_{ngeq 1}H_n z^n$ and $frac{1}{4^n}binom{2n}{n}=frac{2}{pi}int_{0}^{pi/2}left(costhetaright)^{2n},dtheta$, multiply both sides by $(-1)^k binom{k}{h}$, sum over $h=0,1,ldots,k$ and finish by invoking Fubini's theorem (allowing to switch the integrals with respect to $dtheta$ and $dz$) and the Fourier series $sum_{mgeq 1}frac{cos(mvarphi)}{m}$ and $sum_{mgeq 1}frac{sin(mvarphi)}{m}$.



        The only obstruction is that $frac{1}{2n-2h-1}=int_{0}^{1}z^nleft[frac{1}{2z^{h+3/2}}right],dz$ does not hold unconditionally: we would have been happier in having rising Pochhammer symbols rather than falling ones. On the other hand, reindexing fixes this issue. Since $binom{2n+2}{n+1} = frac{2(2n+1)}{n+1}binom{2n}{n}$, the original series can be written as



        $$ sum_{ngeq 1}frac{H_n binom{2n}{n}}{4^n(2n-1)} = sum_{ngeq 0}frac{2H_{n+1}binom{2n}{n}}{4^{n+1}(n+1)}=-frac{1}{pi}int_{0}^{1}sum_{ngeq 0}int_{0}^{pi/2}z^nleft(costhetaright)^{2n}log(1-z),dtheta,dz $$
        or
        $$ -frac{1}{pi}int_{0}^{1}int_{0}^{pi/2}frac{log(1-z)}{1-zcos^2theta},dtheta,dz =-frac{1}{2}int_{0}^{1}frac{log(1-z)}{sqrt{1-z}},dz,$$
        clearly given by a derivative of the Beta function. This approach works also by replacing $(2n-1)$ with $(2n-1)cdots(2n-2k-1)$, you just have to be careful in managing the involved constants depending on $k$.



        On second thought, I have just applied the binomial transform in disguise.






        share|cite|improve this answer











        $endgroup$



        What about reindexing and induction? The terms $frac{1}{(2n-1)cdots(2n-2k-1)}$ have a nice telescopic structure: by the residue theorem
        $$ frac{1}{(2n-1)(2n-3)cdots(2n-2k-1)}=(-1)^ksum_{h=0}^{k}frac{(-1)^h}{(2n-2h-1)}cdotfrac{1}{2^{k+1}(2h)!!(2k-2h)!!} $$
        equals
        $$ frac{(-1)^k}{2^{2k+1}k!} sum_{h=0}^{k}frac{(-1)^h}{(2n-2h-1)}binom{k}{h}.$$
        The natural temptation is now to compute
        $$ sum_{ngeq 1}frac{H_n}{4^{n}}binom{2n}{n}frac{1}{2n-2h-1}$$
        through $frac{-log(1-z)}{1-z}=sum_{ngeq 1}H_n z^n$ and $frac{1}{4^n}binom{2n}{n}=frac{2}{pi}int_{0}^{pi/2}left(costhetaright)^{2n},dtheta$, multiply both sides by $(-1)^k binom{k}{h}$, sum over $h=0,1,ldots,k$ and finish by invoking Fubini's theorem (allowing to switch the integrals with respect to $dtheta$ and $dz$) and the Fourier series $sum_{mgeq 1}frac{cos(mvarphi)}{m}$ and $sum_{mgeq 1}frac{sin(mvarphi)}{m}$.



        The only obstruction is that $frac{1}{2n-2h-1}=int_{0}^{1}z^nleft[frac{1}{2z^{h+3/2}}right],dz$ does not hold unconditionally: we would have been happier in having rising Pochhammer symbols rather than falling ones. On the other hand, reindexing fixes this issue. Since $binom{2n+2}{n+1} = frac{2(2n+1)}{n+1}binom{2n}{n}$, the original series can be written as



        $$ sum_{ngeq 1}frac{H_n binom{2n}{n}}{4^n(2n-1)} = sum_{ngeq 0}frac{2H_{n+1}binom{2n}{n}}{4^{n+1}(n+1)}=-frac{1}{pi}int_{0}^{1}sum_{ngeq 0}int_{0}^{pi/2}z^nleft(costhetaright)^{2n}log(1-z),dtheta,dz $$
        or
        $$ -frac{1}{pi}int_{0}^{1}int_{0}^{pi/2}frac{log(1-z)}{1-zcos^2theta},dtheta,dz =-frac{1}{2}int_{0}^{1}frac{log(1-z)}{sqrt{1-z}},dz,$$
        clearly given by a derivative of the Beta function. This approach works also by replacing $(2n-1)$ with $(2n-1)cdots(2n-2k-1)$, you just have to be careful in managing the involved constants depending on $k$.



        On second thought, I have just applied the binomial transform in disguise.







        share|cite|improve this answer














        share|cite|improve this answer



        share|cite|improve this answer








        edited Jan 2 at 21:21

























        answered Jan 2 at 21:14









        Jack D'AurizioJack D'Aurizio

        289k33281661




        289k33281661






























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