A Series For the Golden Ratio












4















Question: Can we show that $$phi=frac{1}{2}+frac{11}{2}sum_{n=0}^inftyfrac{(2n)!}{5^{3n+1}(n!)^2} $$; where $phi={1+sqrt{5} above 1.5pt 2}$ is the golden ratio ?





The work here is showing that $sum_{n=0}^inftyfrac{(2n)!}{5^{3n+1}(n!)^2} =frac{sqrt{5}}{11};$ which is a play on a similar series involving the square root of $2.$






calculus sequences-and-series golden-ratio







share|cite|improve this question




















  • 7




    $$sum_{ngeq 0}frac{binom{2n}{n}}{4^n}x^{n} = frac{1}{sqrt{1-x}}$$ (for $-1leq x < 1$) is everything you need.
    – Jack D'Aurizio
    2 hours ago










  • I am pretty sure I have seen this before on MSE. It was a post that asked if this series is true and it was painted on something like a poster, or on a book.
    – Zacky
    2 hours ago












  • Found the one I was thinking of: math.stackexchange.com/q/1470099/515527. Well I was wrong :D
    – Zacky
    2 hours ago












  • @Zacky : the comment by robjon in the post you found is relevant - namely the generating function of the central binomial coefficients.
    – Antonio Hernandez Maquivar
    2 hours ago












  • Any reason why this question is being voted to close. Apparently someone believes it is off-topic and not about mathematics ?
    – Antonio Hernandez Maquivar
    2 mins ago
















4















Question: Can we show that $$phi=frac{1}{2}+frac{11}{2}sum_{n=0}^inftyfrac{(2n)!}{5^{3n+1}(n!)^2} $$; where $phi={1+sqrt{5} above 1.5pt 2}$ is the golden ratio ?





The work here is showing that $sum_{n=0}^inftyfrac{(2n)!}{5^{3n+1}(n!)^2} =frac{sqrt{5}}{11};$ which is a play on a similar series involving the square root of $2.$






calculus sequences-and-series golden-ratio







share|cite|improve this question




















  • 7




    $$sum_{ngeq 0}frac{binom{2n}{n}}{4^n}x^{n} = frac{1}{sqrt{1-x}}$$ (for $-1leq x < 1$) is everything you need.
    – Jack D'Aurizio
    2 hours ago










  • I am pretty sure I have seen this before on MSE. It was a post that asked if this series is true and it was painted on something like a poster, or on a book.
    – Zacky
    2 hours ago












  • Found the one I was thinking of: math.stackexchange.com/q/1470099/515527. Well I was wrong :D
    – Zacky
    2 hours ago












  • @Zacky : the comment by robjon in the post you found is relevant - namely the generating function of the central binomial coefficients.
    – Antonio Hernandez Maquivar
    2 hours ago












  • Any reason why this question is being voted to close. Apparently someone believes it is off-topic and not about mathematics ?
    – Antonio Hernandez Maquivar
    2 mins ago














4












4








4


1






Question: Can we show that $$phi=frac{1}{2}+frac{11}{2}sum_{n=0}^inftyfrac{(2n)!}{5^{3n+1}(n!)^2} $$; where $phi={1+sqrt{5} above 1.5pt 2}$ is the golden ratio ?





The work here is showing that $sum_{n=0}^inftyfrac{(2n)!}{5^{3n+1}(n!)^2} =frac{sqrt{5}}{11};$ which is a play on a similar series involving the square root of $2.$






calculus sequences-and-series golden-ratio







share|cite|improve this question
















Question: Can we show that $$phi=frac{1}{2}+frac{11}{2}sum_{n=0}^inftyfrac{(2n)!}{5^{3n+1}(n!)^2} $$; where $phi={1+sqrt{5} above 1.5pt 2}$ is the golden ratio ?





The work here is showing that $sum_{n=0}^inftyfrac{(2n)!}{5^{3n+1}(n!)^2} =frac{sqrt{5}}{11};$ which is a play on a similar series involving the square root of $2.$






calculus sequences-and-series golden-ratio




calculus sequences-and-series golden-ratio






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited 3 mins ago

























asked 2 hours ago









Antonio Hernandez Maquivar

1,363619




1,363619








  • 7




    $$sum_{ngeq 0}frac{binom{2n}{n}}{4^n}x^{n} = frac{1}{sqrt{1-x}}$$ (for $-1leq x < 1$) is everything you need.
    – Jack D'Aurizio
    2 hours ago










  • I am pretty sure I have seen this before on MSE. It was a post that asked if this series is true and it was painted on something like a poster, or on a book.
    – Zacky
    2 hours ago












  • Found the one I was thinking of: math.stackexchange.com/q/1470099/515527. Well I was wrong :D
    – Zacky
    2 hours ago












  • @Zacky : the comment by robjon in the post you found is relevant - namely the generating function of the central binomial coefficients.
    – Antonio Hernandez Maquivar
    2 hours ago












  • Any reason why this question is being voted to close. Apparently someone believes it is off-topic and not about mathematics ?
    – Antonio Hernandez Maquivar
    2 mins ago














  • 7




    $$sum_{ngeq 0}frac{binom{2n}{n}}{4^n}x^{n} = frac{1}{sqrt{1-x}}$$ (for $-1leq x < 1$) is everything you need.
    – Jack D'Aurizio
    2 hours ago










  • I am pretty sure I have seen this before on MSE. It was a post that asked if this series is true and it was painted on something like a poster, or on a book.
    – Zacky
    2 hours ago












  • Found the one I was thinking of: math.stackexchange.com/q/1470099/515527. Well I was wrong :D
    – Zacky
    2 hours ago












  • @Zacky : the comment by robjon in the post you found is relevant - namely the generating function of the central binomial coefficients.
    – Antonio Hernandez Maquivar
    2 hours ago












  • Any reason why this question is being voted to close. Apparently someone believes it is off-topic and not about mathematics ?
    – Antonio Hernandez Maquivar
    2 mins ago








7




7




$$sum_{ngeq 0}frac{binom{2n}{n}}{4^n}x^{n} = frac{1}{sqrt{1-x}}$$ (for $-1leq x < 1$) is everything you need.
– Jack D'Aurizio
2 hours ago




$$sum_{ngeq 0}frac{binom{2n}{n}}{4^n}x^{n} = frac{1}{sqrt{1-x}}$$ (for $-1leq x < 1$) is everything you need.
– Jack D'Aurizio
2 hours ago












I am pretty sure I have seen this before on MSE. It was a post that asked if this series is true and it was painted on something like a poster, or on a book.
– Zacky
2 hours ago






I am pretty sure I have seen this before on MSE. It was a post that asked if this series is true and it was painted on something like a poster, or on a book.
– Zacky
2 hours ago














Found the one I was thinking of: math.stackexchange.com/q/1470099/515527. Well I was wrong :D
– Zacky
2 hours ago






Found the one I was thinking of: math.stackexchange.com/q/1470099/515527. Well I was wrong :D
– Zacky
2 hours ago














@Zacky : the comment by robjon in the post you found is relevant - namely the generating function of the central binomial coefficients.
– Antonio Hernandez Maquivar
2 hours ago






@Zacky : the comment by robjon in the post you found is relevant - namely the generating function of the central binomial coefficients.
– Antonio Hernandez Maquivar
2 hours ago














Any reason why this question is being voted to close. Apparently someone believes it is off-topic and not about mathematics ?
– Antonio Hernandez Maquivar
2 mins ago




Any reason why this question is being voted to close. Apparently someone believes it is off-topic and not about mathematics ?
– Antonio Hernandez Maquivar
2 mins ago










2 Answers
2






active

oldest

votes


















7














Note that




$$frac1{sqrt{1-4x}}=sum_{n=0}^{infty}binom{2n}n x^n$$




Lets rewrite your sum as the following to get



$$sum_{n=0}^inftyfrac{(2n)!}{5^{3n+1}(n!)^2}=frac15sum_{n=0}^inftybinom{2n}nleft(frac1{5^3}right)^n=frac15frac1{sqrt{1-left(frac4{5^3}right)}}=frac{sqrt 5}{11}$$



And therefore you can correctly conclude that



$$frac{1}{2}+frac{11}{2}sum_{n=0}^inftyfrac{(2n)!}{5^{3n+1}(n!)^2}=frac12+frac{11}2frac{sqrt 5}{11}=frac{1+sqrt 5}2$$




$$therefore~frac{1}{2}+frac{11}{2}sum_{n=0}^inftyfrac{(2n)!}{5^{3n+1}(n!)^2}=phi$$







share|cite|improve this answer





















  • What is the motivation behind: $frac1{sqrt{1-4x}}=sum_{n=0}^{infty}binom{2n}n x^n $ ?
    – Antonio Hernandez Maquivar
    2 hours ago










  • @AntonioHernandezMaquivar Do you mean how to derive this representation? Or what dou you mean?
    – mrtaurho
    2 hours ago












  • Why should I have thought of $frac1{sqrt{1-4x}}=sum_{n=0}^{infty}binom{2n}n x^n$ as a starting point to prove my own series ?
    – Antonio Hernandez Maquivar
    2 hours ago










  • @AntonioHernandezMaquivar: $$frac{d^n}{dx^n}left.left(frac{1}{sqrt{1-x}}right)right|_{x=0}$$ is straightforward to compute by induction or by invoking the extended binomial theorem.
    – Jack D'Aurizio
    2 hours ago








  • 1




    @AntonioHernandezMaquivar Since $frac{(2n)!}{(n!)^2}=binom{2n}n$ which is called the central binomial coefficient. This one is used quite often for the purpose of expressing series expansions of well-know functions such as the $arcsin$ or square root function. Thus, seeing this structure within a given series implies a possible connection to one of these functions which I conjectured and it turned out that the series expansion of the function $frac1{sqrt{1-4x}}$ is a good choice in this situtation.
    – mrtaurho
    2 hours ago



















3














Using Calculating $1+frac13+frac{1cdot3}{3cdot6}+frac{1cdot3cdot5}{3cdot6cdot9}+frac{1cdot3cdot5cdot7}{3cdot6cdot9cdot12}+dots? $,



$$dfrac{(2n)!}{5^{3n+1}(n!)^2}=dfrac{2^ncdot1cdot3cdot5cdots(2n-3)(2n-1)}{5^{3n+1}n!}=dfrac{-dfrac12left(-dfrac12-1right)cdotsleft(-dfrac12-(n-1)right)}{n!cdot5}left(-dfrac4{5^3}right)^n$$



$$implies5sum_{n=0}^inftydfrac{(2n)!}{5^{3n+1}(n!)^2}=left(1-dfrac4{5^3}right)^{-1/2}=?$$



Alternatively using Calculating $1+frac13+frac{1cdot3}{3cdot6}+frac{1cdot3cdot5}{3cdot6cdot9}+frac{1cdot3cdot5cdot7}{3cdot6cdot9cdot12}+dots? $,



Using Generalized Binomial Expansion, $$(1+x)^m=1+mx+frac{m(m-1)}{2!}x^2+frac{m(m-1)(m-2)}{3!}x^3+cdots$$ given the converge holds



$mx=dfrac{2!}{5^3(1!)^2},dfrac{m(m-1)}2x^2=dfrac{4!}{5^6(2!)^2}$



$implies m=-dfrac12,x=-dfrac4{5^3}$






share|cite|improve this answer





















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    2 Answers
    2






    active

    oldest

    votes








    2 Answers
    2






    active

    oldest

    votes









    active

    oldest

    votes






    active

    oldest

    votes









    7














    Note that




    $$frac1{sqrt{1-4x}}=sum_{n=0}^{infty}binom{2n}n x^n$$




    Lets rewrite your sum as the following to get



    $$sum_{n=0}^inftyfrac{(2n)!}{5^{3n+1}(n!)^2}=frac15sum_{n=0}^inftybinom{2n}nleft(frac1{5^3}right)^n=frac15frac1{sqrt{1-left(frac4{5^3}right)}}=frac{sqrt 5}{11}$$



    And therefore you can correctly conclude that



    $$frac{1}{2}+frac{11}{2}sum_{n=0}^inftyfrac{(2n)!}{5^{3n+1}(n!)^2}=frac12+frac{11}2frac{sqrt 5}{11}=frac{1+sqrt 5}2$$




    $$therefore~frac{1}{2}+frac{11}{2}sum_{n=0}^inftyfrac{(2n)!}{5^{3n+1}(n!)^2}=phi$$







    share|cite|improve this answer





















    • What is the motivation behind: $frac1{sqrt{1-4x}}=sum_{n=0}^{infty}binom{2n}n x^n $ ?
      – Antonio Hernandez Maquivar
      2 hours ago










    • @AntonioHernandezMaquivar Do you mean how to derive this representation? Or what dou you mean?
      – mrtaurho
      2 hours ago












    • Why should I have thought of $frac1{sqrt{1-4x}}=sum_{n=0}^{infty}binom{2n}n x^n$ as a starting point to prove my own series ?
      – Antonio Hernandez Maquivar
      2 hours ago










    • @AntonioHernandezMaquivar: $$frac{d^n}{dx^n}left.left(frac{1}{sqrt{1-x}}right)right|_{x=0}$$ is straightforward to compute by induction or by invoking the extended binomial theorem.
      – Jack D'Aurizio
      2 hours ago








    • 1




      @AntonioHernandezMaquivar Since $frac{(2n)!}{(n!)^2}=binom{2n}n$ which is called the central binomial coefficient. This one is used quite often for the purpose of expressing series expansions of well-know functions such as the $arcsin$ or square root function. Thus, seeing this structure within a given series implies a possible connection to one of these functions which I conjectured and it turned out that the series expansion of the function $frac1{sqrt{1-4x}}$ is a good choice in this situtation.
      – mrtaurho
      2 hours ago
















    7














    Note that




    $$frac1{sqrt{1-4x}}=sum_{n=0}^{infty}binom{2n}n x^n$$




    Lets rewrite your sum as the following to get



    $$sum_{n=0}^inftyfrac{(2n)!}{5^{3n+1}(n!)^2}=frac15sum_{n=0}^inftybinom{2n}nleft(frac1{5^3}right)^n=frac15frac1{sqrt{1-left(frac4{5^3}right)}}=frac{sqrt 5}{11}$$



    And therefore you can correctly conclude that



    $$frac{1}{2}+frac{11}{2}sum_{n=0}^inftyfrac{(2n)!}{5^{3n+1}(n!)^2}=frac12+frac{11}2frac{sqrt 5}{11}=frac{1+sqrt 5}2$$




    $$therefore~frac{1}{2}+frac{11}{2}sum_{n=0}^inftyfrac{(2n)!}{5^{3n+1}(n!)^2}=phi$$







    share|cite|improve this answer





















    • What is the motivation behind: $frac1{sqrt{1-4x}}=sum_{n=0}^{infty}binom{2n}n x^n $ ?
      – Antonio Hernandez Maquivar
      2 hours ago










    • @AntonioHernandezMaquivar Do you mean how to derive this representation? Or what dou you mean?
      – mrtaurho
      2 hours ago












    • Why should I have thought of $frac1{sqrt{1-4x}}=sum_{n=0}^{infty}binom{2n}n x^n$ as a starting point to prove my own series ?
      – Antonio Hernandez Maquivar
      2 hours ago










    • @AntonioHernandezMaquivar: $$frac{d^n}{dx^n}left.left(frac{1}{sqrt{1-x}}right)right|_{x=0}$$ is straightforward to compute by induction or by invoking the extended binomial theorem.
      – Jack D'Aurizio
      2 hours ago








    • 1




      @AntonioHernandezMaquivar Since $frac{(2n)!}{(n!)^2}=binom{2n}n$ which is called the central binomial coefficient. This one is used quite often for the purpose of expressing series expansions of well-know functions such as the $arcsin$ or square root function. Thus, seeing this structure within a given series implies a possible connection to one of these functions which I conjectured and it turned out that the series expansion of the function $frac1{sqrt{1-4x}}$ is a good choice in this situtation.
      – mrtaurho
      2 hours ago














    7












    7








    7






    Note that




    $$frac1{sqrt{1-4x}}=sum_{n=0}^{infty}binom{2n}n x^n$$




    Lets rewrite your sum as the following to get



    $$sum_{n=0}^inftyfrac{(2n)!}{5^{3n+1}(n!)^2}=frac15sum_{n=0}^inftybinom{2n}nleft(frac1{5^3}right)^n=frac15frac1{sqrt{1-left(frac4{5^3}right)}}=frac{sqrt 5}{11}$$



    And therefore you can correctly conclude that



    $$frac{1}{2}+frac{11}{2}sum_{n=0}^inftyfrac{(2n)!}{5^{3n+1}(n!)^2}=frac12+frac{11}2frac{sqrt 5}{11}=frac{1+sqrt 5}2$$




    $$therefore~frac{1}{2}+frac{11}{2}sum_{n=0}^inftyfrac{(2n)!}{5^{3n+1}(n!)^2}=phi$$







    share|cite|improve this answer












    Note that




    $$frac1{sqrt{1-4x}}=sum_{n=0}^{infty}binom{2n}n x^n$$




    Lets rewrite your sum as the following to get



    $$sum_{n=0}^inftyfrac{(2n)!}{5^{3n+1}(n!)^2}=frac15sum_{n=0}^inftybinom{2n}nleft(frac1{5^3}right)^n=frac15frac1{sqrt{1-left(frac4{5^3}right)}}=frac{sqrt 5}{11}$$



    And therefore you can correctly conclude that



    $$frac{1}{2}+frac{11}{2}sum_{n=0}^inftyfrac{(2n)!}{5^{3n+1}(n!)^2}=frac12+frac{11}2frac{sqrt 5}{11}=frac{1+sqrt 5}2$$




    $$therefore~frac{1}{2}+frac{11}{2}sum_{n=0}^inftyfrac{(2n)!}{5^{3n+1}(n!)^2}=phi$$








    share|cite|improve this answer












    share|cite|improve this answer



    share|cite|improve this answer










    answered 2 hours ago









    mrtaurho

    3,53121032




    3,53121032












    • What is the motivation behind: $frac1{sqrt{1-4x}}=sum_{n=0}^{infty}binom{2n}n x^n $ ?
      – Antonio Hernandez Maquivar
      2 hours ago










    • @AntonioHernandezMaquivar Do you mean how to derive this representation? Or what dou you mean?
      – mrtaurho
      2 hours ago












    • Why should I have thought of $frac1{sqrt{1-4x}}=sum_{n=0}^{infty}binom{2n}n x^n$ as a starting point to prove my own series ?
      – Antonio Hernandez Maquivar
      2 hours ago










    • @AntonioHernandezMaquivar: $$frac{d^n}{dx^n}left.left(frac{1}{sqrt{1-x}}right)right|_{x=0}$$ is straightforward to compute by induction or by invoking the extended binomial theorem.
      – Jack D'Aurizio
      2 hours ago








    • 1




      @AntonioHernandezMaquivar Since $frac{(2n)!}{(n!)^2}=binom{2n}n$ which is called the central binomial coefficient. This one is used quite often for the purpose of expressing series expansions of well-know functions such as the $arcsin$ or square root function. Thus, seeing this structure within a given series implies a possible connection to one of these functions which I conjectured and it turned out that the series expansion of the function $frac1{sqrt{1-4x}}$ is a good choice in this situtation.
      – mrtaurho
      2 hours ago


















    • What is the motivation behind: $frac1{sqrt{1-4x}}=sum_{n=0}^{infty}binom{2n}n x^n $ ?
      – Antonio Hernandez Maquivar
      2 hours ago










    • @AntonioHernandezMaquivar Do you mean how to derive this representation? Or what dou you mean?
      – mrtaurho
      2 hours ago












    • Why should I have thought of $frac1{sqrt{1-4x}}=sum_{n=0}^{infty}binom{2n}n x^n$ as a starting point to prove my own series ?
      – Antonio Hernandez Maquivar
      2 hours ago










    • @AntonioHernandezMaquivar: $$frac{d^n}{dx^n}left.left(frac{1}{sqrt{1-x}}right)right|_{x=0}$$ is straightforward to compute by induction or by invoking the extended binomial theorem.
      – Jack D'Aurizio
      2 hours ago








    • 1




      @AntonioHernandezMaquivar Since $frac{(2n)!}{(n!)^2}=binom{2n}n$ which is called the central binomial coefficient. This one is used quite often for the purpose of expressing series expansions of well-know functions such as the $arcsin$ or square root function. Thus, seeing this structure within a given series implies a possible connection to one of these functions which I conjectured and it turned out that the series expansion of the function $frac1{sqrt{1-4x}}$ is a good choice in this situtation.
      – mrtaurho
      2 hours ago
















    What is the motivation behind: $frac1{sqrt{1-4x}}=sum_{n=0}^{infty}binom{2n}n x^n $ ?
    – Antonio Hernandez Maquivar
    2 hours ago




    What is the motivation behind: $frac1{sqrt{1-4x}}=sum_{n=0}^{infty}binom{2n}n x^n $ ?
    – Antonio Hernandez Maquivar
    2 hours ago












    @AntonioHernandezMaquivar Do you mean how to derive this representation? Or what dou you mean?
    – mrtaurho
    2 hours ago






    @AntonioHernandezMaquivar Do you mean how to derive this representation? Or what dou you mean?
    – mrtaurho
    2 hours ago














    Why should I have thought of $frac1{sqrt{1-4x}}=sum_{n=0}^{infty}binom{2n}n x^n$ as a starting point to prove my own series ?
    – Antonio Hernandez Maquivar
    2 hours ago




    Why should I have thought of $frac1{sqrt{1-4x}}=sum_{n=0}^{infty}binom{2n}n x^n$ as a starting point to prove my own series ?
    – Antonio Hernandez Maquivar
    2 hours ago












    @AntonioHernandezMaquivar: $$frac{d^n}{dx^n}left.left(frac{1}{sqrt{1-x}}right)right|_{x=0}$$ is straightforward to compute by induction or by invoking the extended binomial theorem.
    – Jack D'Aurizio
    2 hours ago






    @AntonioHernandezMaquivar: $$frac{d^n}{dx^n}left.left(frac{1}{sqrt{1-x}}right)right|_{x=0}$$ is straightforward to compute by induction or by invoking the extended binomial theorem.
    – Jack D'Aurizio
    2 hours ago






    1




    1




    @AntonioHernandezMaquivar Since $frac{(2n)!}{(n!)^2}=binom{2n}n$ which is called the central binomial coefficient. This one is used quite often for the purpose of expressing series expansions of well-know functions such as the $arcsin$ or square root function. Thus, seeing this structure within a given series implies a possible connection to one of these functions which I conjectured and it turned out that the series expansion of the function $frac1{sqrt{1-4x}}$ is a good choice in this situtation.
    – mrtaurho
    2 hours ago




    @AntonioHernandezMaquivar Since $frac{(2n)!}{(n!)^2}=binom{2n}n$ which is called the central binomial coefficient. This one is used quite often for the purpose of expressing series expansions of well-know functions such as the $arcsin$ or square root function. Thus, seeing this structure within a given series implies a possible connection to one of these functions which I conjectured and it turned out that the series expansion of the function $frac1{sqrt{1-4x}}$ is a good choice in this situtation.
    – mrtaurho
    2 hours ago











    3














    Using Calculating $1+frac13+frac{1cdot3}{3cdot6}+frac{1cdot3cdot5}{3cdot6cdot9}+frac{1cdot3cdot5cdot7}{3cdot6cdot9cdot12}+dots? $,



    $$dfrac{(2n)!}{5^{3n+1}(n!)^2}=dfrac{2^ncdot1cdot3cdot5cdots(2n-3)(2n-1)}{5^{3n+1}n!}=dfrac{-dfrac12left(-dfrac12-1right)cdotsleft(-dfrac12-(n-1)right)}{n!cdot5}left(-dfrac4{5^3}right)^n$$



    $$implies5sum_{n=0}^inftydfrac{(2n)!}{5^{3n+1}(n!)^2}=left(1-dfrac4{5^3}right)^{-1/2}=?$$



    Alternatively using Calculating $1+frac13+frac{1cdot3}{3cdot6}+frac{1cdot3cdot5}{3cdot6cdot9}+frac{1cdot3cdot5cdot7}{3cdot6cdot9cdot12}+dots? $,



    Using Generalized Binomial Expansion, $$(1+x)^m=1+mx+frac{m(m-1)}{2!}x^2+frac{m(m-1)(m-2)}{3!}x^3+cdots$$ given the converge holds



    $mx=dfrac{2!}{5^3(1!)^2},dfrac{m(m-1)}2x^2=dfrac{4!}{5^6(2!)^2}$



    $implies m=-dfrac12,x=-dfrac4{5^3}$






    share|cite|improve this answer


























      3














      Using Calculating $1+frac13+frac{1cdot3}{3cdot6}+frac{1cdot3cdot5}{3cdot6cdot9}+frac{1cdot3cdot5cdot7}{3cdot6cdot9cdot12}+dots? $,



      $$dfrac{(2n)!}{5^{3n+1}(n!)^2}=dfrac{2^ncdot1cdot3cdot5cdots(2n-3)(2n-1)}{5^{3n+1}n!}=dfrac{-dfrac12left(-dfrac12-1right)cdotsleft(-dfrac12-(n-1)right)}{n!cdot5}left(-dfrac4{5^3}right)^n$$



      $$implies5sum_{n=0}^inftydfrac{(2n)!}{5^{3n+1}(n!)^2}=left(1-dfrac4{5^3}right)^{-1/2}=?$$



      Alternatively using Calculating $1+frac13+frac{1cdot3}{3cdot6}+frac{1cdot3cdot5}{3cdot6cdot9}+frac{1cdot3cdot5cdot7}{3cdot6cdot9cdot12}+dots? $,



      Using Generalized Binomial Expansion, $$(1+x)^m=1+mx+frac{m(m-1)}{2!}x^2+frac{m(m-1)(m-2)}{3!}x^3+cdots$$ given the converge holds



      $mx=dfrac{2!}{5^3(1!)^2},dfrac{m(m-1)}2x^2=dfrac{4!}{5^6(2!)^2}$



      $implies m=-dfrac12,x=-dfrac4{5^3}$






      share|cite|improve this answer
























        3












        3








        3






        Using Calculating $1+frac13+frac{1cdot3}{3cdot6}+frac{1cdot3cdot5}{3cdot6cdot9}+frac{1cdot3cdot5cdot7}{3cdot6cdot9cdot12}+dots? $,



        $$dfrac{(2n)!}{5^{3n+1}(n!)^2}=dfrac{2^ncdot1cdot3cdot5cdots(2n-3)(2n-1)}{5^{3n+1}n!}=dfrac{-dfrac12left(-dfrac12-1right)cdotsleft(-dfrac12-(n-1)right)}{n!cdot5}left(-dfrac4{5^3}right)^n$$



        $$implies5sum_{n=0}^inftydfrac{(2n)!}{5^{3n+1}(n!)^2}=left(1-dfrac4{5^3}right)^{-1/2}=?$$



        Alternatively using Calculating $1+frac13+frac{1cdot3}{3cdot6}+frac{1cdot3cdot5}{3cdot6cdot9}+frac{1cdot3cdot5cdot7}{3cdot6cdot9cdot12}+dots? $,



        Using Generalized Binomial Expansion, $$(1+x)^m=1+mx+frac{m(m-1)}{2!}x^2+frac{m(m-1)(m-2)}{3!}x^3+cdots$$ given the converge holds



        $mx=dfrac{2!}{5^3(1!)^2},dfrac{m(m-1)}2x^2=dfrac{4!}{5^6(2!)^2}$



        $implies m=-dfrac12,x=-dfrac4{5^3}$






        share|cite|improve this answer












        Using Calculating $1+frac13+frac{1cdot3}{3cdot6}+frac{1cdot3cdot5}{3cdot6cdot9}+frac{1cdot3cdot5cdot7}{3cdot6cdot9cdot12}+dots? $,



        $$dfrac{(2n)!}{5^{3n+1}(n!)^2}=dfrac{2^ncdot1cdot3cdot5cdots(2n-3)(2n-1)}{5^{3n+1}n!}=dfrac{-dfrac12left(-dfrac12-1right)cdotsleft(-dfrac12-(n-1)right)}{n!cdot5}left(-dfrac4{5^3}right)^n$$



        $$implies5sum_{n=0}^inftydfrac{(2n)!}{5^{3n+1}(n!)^2}=left(1-dfrac4{5^3}right)^{-1/2}=?$$



        Alternatively using Calculating $1+frac13+frac{1cdot3}{3cdot6}+frac{1cdot3cdot5}{3cdot6cdot9}+frac{1cdot3cdot5cdot7}{3cdot6cdot9cdot12}+dots? $,



        Using Generalized Binomial Expansion, $$(1+x)^m=1+mx+frac{m(m-1)}{2!}x^2+frac{m(m-1)(m-2)}{3!}x^3+cdots$$ given the converge holds



        $mx=dfrac{2!}{5^3(1!)^2},dfrac{m(m-1)}2x^2=dfrac{4!}{5^6(2!)^2}$



        $implies m=-dfrac12,x=-dfrac4{5^3}$







        share|cite|improve this answer












        share|cite|improve this answer



        share|cite|improve this answer










        answered 1 hour ago









        lab bhattacharjee

        223k15156274




        223k15156274






























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