series $sum_{n=1}^infty sum_{m=1}^infty frac{(m+n-1)!}{m!(n-1)!n!(m-1)!}a^m b^n.$












2












$begingroup$


Can anyone please help me with the computation of following series:



$$sum_{n=1}^infty sum_{m=1}^infty frac{(m+n-1)!}{m!(n-1)!n!(m-1)!}a^m b^n.$$



My thoughts: Since $$displaystyle frac{(m+n-1)!}{m!(n-1)!n!(m-1)!} = frac{binom{m+n-1}{m}binom{m+n-1}{m-1}}{(m+n-1)!},$$ by some arrangement this may be the probability of a hypergeometric distribution.



Any help would be appreciated, thanks!






combinatorics algebra-precalculus binomial-theorem







share|cite|improve this question











$endgroup$












  • $begingroup$
    There is an explicit solution (involving the Bessel function $I_1$) in the case $a=b$. I am not so sure about the general case, which boils down to evaluating $$ int_{0}^{+infty} I_1(2sqrt{az})I_1(2sqrt{bz})e^{-z},dz. $$ Of course we may use the asymptotics for Bessel functions to derive accurate approximations.
    $endgroup$
    – Jack D'Aurizio
    Jan 26 at 1:45
















2












$begingroup$


Can anyone please help me with the computation of following series:



$$sum_{n=1}^infty sum_{m=1}^infty frac{(m+n-1)!}{m!(n-1)!n!(m-1)!}a^m b^n.$$



My thoughts: Since $$displaystyle frac{(m+n-1)!}{m!(n-1)!n!(m-1)!} = frac{binom{m+n-1}{m}binom{m+n-1}{m-1}}{(m+n-1)!},$$ by some arrangement this may be the probability of a hypergeometric distribution.



Any help would be appreciated, thanks!






combinatorics algebra-precalculus binomial-theorem







share|cite|improve this question











$endgroup$












  • $begingroup$
    There is an explicit solution (involving the Bessel function $I_1$) in the case $a=b$. I am not so sure about the general case, which boils down to evaluating $$ int_{0}^{+infty} I_1(2sqrt{az})I_1(2sqrt{bz})e^{-z},dz. $$ Of course we may use the asymptotics for Bessel functions to derive accurate approximations.
    $endgroup$
    – Jack D'Aurizio
    Jan 26 at 1:45














2












2








2


1



$begingroup$


Can anyone please help me with the computation of following series:



$$sum_{n=1}^infty sum_{m=1}^infty frac{(m+n-1)!}{m!(n-1)!n!(m-1)!}a^m b^n.$$



My thoughts: Since $$displaystyle frac{(m+n-1)!}{m!(n-1)!n!(m-1)!} = frac{binom{m+n-1}{m}binom{m+n-1}{m-1}}{(m+n-1)!},$$ by some arrangement this may be the probability of a hypergeometric distribution.



Any help would be appreciated, thanks!






combinatorics algebra-precalculus binomial-theorem







share|cite|improve this question











$endgroup$




Can anyone please help me with the computation of following series:



$$sum_{n=1}^infty sum_{m=1}^infty frac{(m+n-1)!}{m!(n-1)!n!(m-1)!}a^m b^n.$$



My thoughts: Since $$displaystyle frac{(m+n-1)!}{m!(n-1)!n!(m-1)!} = frac{binom{m+n-1}{m}binom{m+n-1}{m-1}}{(m+n-1)!},$$ by some arrangement this may be the probability of a hypergeometric distribution.



Any help would be appreciated, thanks!






combinatorics algebra-precalculus binomial-theorem




combinatorics algebra-precalculus binomial-theorem






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Jan 26 at 2:56









Arjang

5,64062364




5,64062364










asked Jan 25 at 16:37









Violet. WViolet. W

313




313












  • $begingroup$
    There is an explicit solution (involving the Bessel function $I_1$) in the case $a=b$. I am not so sure about the general case, which boils down to evaluating $$ int_{0}^{+infty} I_1(2sqrt{az})I_1(2sqrt{bz})e^{-z},dz. $$ Of course we may use the asymptotics for Bessel functions to derive accurate approximations.
    $endgroup$
    – Jack D'Aurizio
    Jan 26 at 1:45


















  • $begingroup$
    There is an explicit solution (involving the Bessel function $I_1$) in the case $a=b$. I am not so sure about the general case, which boils down to evaluating $$ int_{0}^{+infty} I_1(2sqrt{az})I_1(2sqrt{bz})e^{-z},dz. $$ Of course we may use the asymptotics for Bessel functions to derive accurate approximations.
    $endgroup$
    – Jack D'Aurizio
    Jan 26 at 1:45
















$begingroup$
There is an explicit solution (involving the Bessel function $I_1$) in the case $a=b$. I am not so sure about the general case, which boils down to evaluating $$ int_{0}^{+infty} I_1(2sqrt{az})I_1(2sqrt{bz})e^{-z},dz. $$ Of course we may use the asymptotics for Bessel functions to derive accurate approximations.
$endgroup$
– Jack D'Aurizio
Jan 26 at 1:45




$begingroup$
There is an explicit solution (involving the Bessel function $I_1$) in the case $a=b$. I am not so sure about the general case, which boils down to evaluating $$ int_{0}^{+infty} I_1(2sqrt{az})I_1(2sqrt{bz})e^{-z},dz. $$ Of course we may use the asymptotics for Bessel functions to derive accurate approximations.
$endgroup$
– Jack D'Aurizio
Jan 26 at 1:45










2 Answers
2






active

oldest

votes


















3












$begingroup$

I am going to outline a nasty trick which allows to prove
$$ sum_{m,ngeq 0}frac{(m+n)!}{m!^2 n!^2}a^m b^n = e^{a+b} I_0(2sqrt{ab}) tag{1}$$
and similarly
$$ sum_{m,ngeq 1}frac{(m+n-1)!}{m!(m-1)! n!(n-1)!}a^m b^n = sqrt{ab}, e^{a+b} I_1(2sqrt{ab})tag{2}$$
We apply two Laplace transforms: the first transform maps the $a$-variable into the $s$-variable and the second transform maps the $b$-variable into the $t$-variable. The double-Laplace-transform of the LHS of $(2)$ equals
$$ frac{1}{(s+t-st)^2}=frac{1}{left[1-(1-s)(1-t)right]^2}=sum_{hgeq 0}frac{(h+1)}{(1-s)^{h+2} (1-t)^{h+2}}tag{3} $$
and by applying the double-inverse-trasform (should it be the inverse-double-transform? I am not really sure) to the terms appearing in the RHS of $(3)$ we get that the LHS of $(2)$ equals
$$ e^{a+b}sum_{hgeq 0}frac{(-a)^{h+1}(-b)^{h+1}(h+1)}{Gamma(h+2)^2} = sqrt{ab}, e^{a+b} I_1(2sqrt{ab}).tag{4}$$
If both $a$ and $b$ are $gg 1$ this is approximately
$$ frac{1}{sqrt{4pi}} e^{(sqrt{a}+sqrt{b})^2}left[(ab)^{1/4}-tfrac{3}{16}(ab)^{-1/4}right]. tag{5} $$






share|cite|improve this answer









$endgroup$













  • $begingroup$
    Thank you Jack, your answer is mind-blowing! There is another way to simplify this series when $a,b<1$. I will post it as an optional answer.
    $endgroup$
    – Violet. W
    Jan 29 at 18:43



















0












$begingroup$

One feasible way to simplify it:



since $$sum_{m=1}^infty frac{(m+n-1)!}{m!(m-1)!} a^m = sum_{m=1}^inftyfrac{m(m+1)...(m+n-1)}{m!}a^m,$$



when $a>0$, one can use the formula for factorial moment of Poisson distribution to simplify this series, and same procedure can be applied to the series of n.






share|cite|improve this answer











$endgroup$













    Your Answer





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    2 Answers
    2






    active

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    2 Answers
    2






    active

    oldest

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    active

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    active

    oldest

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    3












    $begingroup$

    I am going to outline a nasty trick which allows to prove
    $$ sum_{m,ngeq 0}frac{(m+n)!}{m!^2 n!^2}a^m b^n = e^{a+b} I_0(2sqrt{ab}) tag{1}$$
    and similarly
    $$ sum_{m,ngeq 1}frac{(m+n-1)!}{m!(m-1)! n!(n-1)!}a^m b^n = sqrt{ab}, e^{a+b} I_1(2sqrt{ab})tag{2}$$
    We apply two Laplace transforms: the first transform maps the $a$-variable into the $s$-variable and the second transform maps the $b$-variable into the $t$-variable. The double-Laplace-transform of the LHS of $(2)$ equals
    $$ frac{1}{(s+t-st)^2}=frac{1}{left[1-(1-s)(1-t)right]^2}=sum_{hgeq 0}frac{(h+1)}{(1-s)^{h+2} (1-t)^{h+2}}tag{3} $$
    and by applying the double-inverse-trasform (should it be the inverse-double-transform? I am not really sure) to the terms appearing in the RHS of $(3)$ we get that the LHS of $(2)$ equals
    $$ e^{a+b}sum_{hgeq 0}frac{(-a)^{h+1}(-b)^{h+1}(h+1)}{Gamma(h+2)^2} = sqrt{ab}, e^{a+b} I_1(2sqrt{ab}).tag{4}$$
    If both $a$ and $b$ are $gg 1$ this is approximately
    $$ frac{1}{sqrt{4pi}} e^{(sqrt{a}+sqrt{b})^2}left[(ab)^{1/4}-tfrac{3}{16}(ab)^{-1/4}right]. tag{5} $$






    share|cite|improve this answer









    $endgroup$













    • $begingroup$
      Thank you Jack, your answer is mind-blowing! There is another way to simplify this series when $a,b<1$. I will post it as an optional answer.
      $endgroup$
      – Violet. W
      Jan 29 at 18:43
















    3












    $begingroup$

    I am going to outline a nasty trick which allows to prove
    $$ sum_{m,ngeq 0}frac{(m+n)!}{m!^2 n!^2}a^m b^n = e^{a+b} I_0(2sqrt{ab}) tag{1}$$
    and similarly
    $$ sum_{m,ngeq 1}frac{(m+n-1)!}{m!(m-1)! n!(n-1)!}a^m b^n = sqrt{ab}, e^{a+b} I_1(2sqrt{ab})tag{2}$$
    We apply two Laplace transforms: the first transform maps the $a$-variable into the $s$-variable and the second transform maps the $b$-variable into the $t$-variable. The double-Laplace-transform of the LHS of $(2)$ equals
    $$ frac{1}{(s+t-st)^2}=frac{1}{left[1-(1-s)(1-t)right]^2}=sum_{hgeq 0}frac{(h+1)}{(1-s)^{h+2} (1-t)^{h+2}}tag{3} $$
    and by applying the double-inverse-trasform (should it be the inverse-double-transform? I am not really sure) to the terms appearing in the RHS of $(3)$ we get that the LHS of $(2)$ equals
    $$ e^{a+b}sum_{hgeq 0}frac{(-a)^{h+1}(-b)^{h+1}(h+1)}{Gamma(h+2)^2} = sqrt{ab}, e^{a+b} I_1(2sqrt{ab}).tag{4}$$
    If both $a$ and $b$ are $gg 1$ this is approximately
    $$ frac{1}{sqrt{4pi}} e^{(sqrt{a}+sqrt{b})^2}left[(ab)^{1/4}-tfrac{3}{16}(ab)^{-1/4}right]. tag{5} $$






    share|cite|improve this answer









    $endgroup$













    • $begingroup$
      Thank you Jack, your answer is mind-blowing! There is another way to simplify this series when $a,b<1$. I will post it as an optional answer.
      $endgroup$
      – Violet. W
      Jan 29 at 18:43














    3












    3








    3





    $begingroup$

    I am going to outline a nasty trick which allows to prove
    $$ sum_{m,ngeq 0}frac{(m+n)!}{m!^2 n!^2}a^m b^n = e^{a+b} I_0(2sqrt{ab}) tag{1}$$
    and similarly
    $$ sum_{m,ngeq 1}frac{(m+n-1)!}{m!(m-1)! n!(n-1)!}a^m b^n = sqrt{ab}, e^{a+b} I_1(2sqrt{ab})tag{2}$$
    We apply two Laplace transforms: the first transform maps the $a$-variable into the $s$-variable and the second transform maps the $b$-variable into the $t$-variable. The double-Laplace-transform of the LHS of $(2)$ equals
    $$ frac{1}{(s+t-st)^2}=frac{1}{left[1-(1-s)(1-t)right]^2}=sum_{hgeq 0}frac{(h+1)}{(1-s)^{h+2} (1-t)^{h+2}}tag{3} $$
    and by applying the double-inverse-trasform (should it be the inverse-double-transform? I am not really sure) to the terms appearing in the RHS of $(3)$ we get that the LHS of $(2)$ equals
    $$ e^{a+b}sum_{hgeq 0}frac{(-a)^{h+1}(-b)^{h+1}(h+1)}{Gamma(h+2)^2} = sqrt{ab}, e^{a+b} I_1(2sqrt{ab}).tag{4}$$
    If both $a$ and $b$ are $gg 1$ this is approximately
    $$ frac{1}{sqrt{4pi}} e^{(sqrt{a}+sqrt{b})^2}left[(ab)^{1/4}-tfrac{3}{16}(ab)^{-1/4}right]. tag{5} $$






    share|cite|improve this answer









    $endgroup$



    I am going to outline a nasty trick which allows to prove
    $$ sum_{m,ngeq 0}frac{(m+n)!}{m!^2 n!^2}a^m b^n = e^{a+b} I_0(2sqrt{ab}) tag{1}$$
    and similarly
    $$ sum_{m,ngeq 1}frac{(m+n-1)!}{m!(m-1)! n!(n-1)!}a^m b^n = sqrt{ab}, e^{a+b} I_1(2sqrt{ab})tag{2}$$
    We apply two Laplace transforms: the first transform maps the $a$-variable into the $s$-variable and the second transform maps the $b$-variable into the $t$-variable. The double-Laplace-transform of the LHS of $(2)$ equals
    $$ frac{1}{(s+t-st)^2}=frac{1}{left[1-(1-s)(1-t)right]^2}=sum_{hgeq 0}frac{(h+1)}{(1-s)^{h+2} (1-t)^{h+2}}tag{3} $$
    and by applying the double-inverse-trasform (should it be the inverse-double-transform? I am not really sure) to the terms appearing in the RHS of $(3)$ we get that the LHS of $(2)$ equals
    $$ e^{a+b}sum_{hgeq 0}frac{(-a)^{h+1}(-b)^{h+1}(h+1)}{Gamma(h+2)^2} = sqrt{ab}, e^{a+b} I_1(2sqrt{ab}).tag{4}$$
    If both $a$ and $b$ are $gg 1$ this is approximately
    $$ frac{1}{sqrt{4pi}} e^{(sqrt{a}+sqrt{b})^2}left[(ab)^{1/4}-tfrac{3}{16}(ab)^{-1/4}right]. tag{5} $$







    share|cite|improve this answer












    share|cite|improve this answer



    share|cite|improve this answer










    answered Jan 26 at 2:46









    Jack D'AurizioJack D'Aurizio

    291k33284667




    291k33284667












    • $begingroup$
      Thank you Jack, your answer is mind-blowing! There is another way to simplify this series when $a,b<1$. I will post it as an optional answer.
      $endgroup$
      – Violet. W
      Jan 29 at 18:43


















    • $begingroup$
      Thank you Jack, your answer is mind-blowing! There is another way to simplify this series when $a,b<1$. I will post it as an optional answer.
      $endgroup$
      – Violet. W
      Jan 29 at 18:43
















    $begingroup$
    Thank you Jack, your answer is mind-blowing! There is another way to simplify this series when $a,b<1$. I will post it as an optional answer.
    $endgroup$
    – Violet. W
    Jan 29 at 18:43




    $begingroup$
    Thank you Jack, your answer is mind-blowing! There is another way to simplify this series when $a,b<1$. I will post it as an optional answer.
    $endgroup$
    – Violet. W
    Jan 29 at 18:43











    0












    $begingroup$

    One feasible way to simplify it:



    since $$sum_{m=1}^infty frac{(m+n-1)!}{m!(m-1)!} a^m = sum_{m=1}^inftyfrac{m(m+1)...(m+n-1)}{m!}a^m,$$



    when $a>0$, one can use the formula for factorial moment of Poisson distribution to simplify this series, and same procedure can be applied to the series of n.






    share|cite|improve this answer











    $endgroup$


















      0












      $begingroup$

      One feasible way to simplify it:



      since $$sum_{m=1}^infty frac{(m+n-1)!}{m!(m-1)!} a^m = sum_{m=1}^inftyfrac{m(m+1)...(m+n-1)}{m!}a^m,$$



      when $a>0$, one can use the formula for factorial moment of Poisson distribution to simplify this series, and same procedure can be applied to the series of n.






      share|cite|improve this answer











      $endgroup$
















        0












        0








        0





        $begingroup$

        One feasible way to simplify it:



        since $$sum_{m=1}^infty frac{(m+n-1)!}{m!(m-1)!} a^m = sum_{m=1}^inftyfrac{m(m+1)...(m+n-1)}{m!}a^m,$$



        when $a>0$, one can use the formula for factorial moment of Poisson distribution to simplify this series, and same procedure can be applied to the series of n.






        share|cite|improve this answer











        $endgroup$



        One feasible way to simplify it:



        since $$sum_{m=1}^infty frac{(m+n-1)!}{m!(m-1)!} a^m = sum_{m=1}^inftyfrac{m(m+1)...(m+n-1)}{m!}a^m,$$



        when $a>0$, one can use the formula for factorial moment of Poisson distribution to simplify this series, and same procedure can be applied to the series of n.







        share|cite|improve this answer














        share|cite|improve this answer



        share|cite|improve this answer








        edited Jan 29 at 19:14

























        answered Jan 29 at 18:47









        Violet. WViolet. W

        313




        313






























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