Evaluate 1 + 3/4 + (3*5)/(4*8)+…












0












$begingroup$


Evaluate
1 + 3/4 + (3.5)/(4.8) + (3.5.7)/(4.8.12) +...



I have simplified it to
Summation ((2k+1)!)/(k!)(k!)) (1/8)^k where k varies from 0 to infinity.



I am not able to to relate this sum with Taylor expansion of some function






real-analysis taylor-expansion







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  • $begingroup$
    math.stackexchange.com/questions/746388/…
    $endgroup$
    – lab bhattacharjee
    Jan 9 at 5:17
















0












$begingroup$


Evaluate
1 + 3/4 + (3.5)/(4.8) + (3.5.7)/(4.8.12) +...



I have simplified it to
Summation ((2k+1)!)/(k!)(k!)) (1/8)^k where k varies from 0 to infinity.



I am not able to to relate this sum with Taylor expansion of some function






real-analysis taylor-expansion







share|cite|improve this question









$endgroup$












  • $begingroup$
    math.stackexchange.com/questions/746388/…
    $endgroup$
    – lab bhattacharjee
    Jan 9 at 5:17














0












0








0





$begingroup$


Evaluate
1 + 3/4 + (3.5)/(4.8) + (3.5.7)/(4.8.12) +...



I have simplified it to
Summation ((2k+1)!)/(k!)(k!)) (1/8)^k where k varies from 0 to infinity.



I am not able to to relate this sum with Taylor expansion of some function






real-analysis taylor-expansion







share|cite|improve this question









$endgroup$




Evaluate
1 + 3/4 + (3.5)/(4.8) + (3.5.7)/(4.8.12) +...



I have simplified it to
Summation ((2k+1)!)/(k!)(k!)) (1/8)^k where k varies from 0 to infinity.



I am not able to to relate this sum with Taylor expansion of some function






real-analysis taylor-expansion




real-analysis taylor-expansion






share|cite|improve this question













share|cite|improve this question











share|cite|improve this question




share|cite|improve this question










asked Jan 9 at 4:45









Swapnil Swapnil

515




515












  • $begingroup$
    math.stackexchange.com/questions/746388/…
    $endgroup$
    – lab bhattacharjee
    Jan 9 at 5:17


















  • $begingroup$
    math.stackexchange.com/questions/746388/…
    $endgroup$
    – lab bhattacharjee
    Jan 9 at 5:17
















$begingroup$
math.stackexchange.com/questions/746388/…
$endgroup$
– lab bhattacharjee
Jan 9 at 5:17




$begingroup$
math.stackexchange.com/questions/746388/…
$endgroup$
– lab bhattacharjee
Jan 9 at 5:17










2 Answers
2






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oldest

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2












$begingroup$

Sounds like you are looking for
$$
sum_{k = 0}^infty (2k+1) binom{2k}{k}left( frac18 right)^k,
$$
so consider differentiating
$$
f(x) = sum_{k = 0}^infty binom{2k}{k} x^{2k+1},
$$
and using $(1/8)^k = (1/sqrt8)^{2k}$






share|cite|improve this answer









$endgroup$





















    0












    $begingroup$

    the series is the binomial expansion of
    $$
    bigg(1 - frac12 bigg)^{-frac32}
    $$



    which gives the sum $2sqrt{2}$






    share|cite|improve this answer











    $endgroup$









    • 1




      $begingroup$
      MathJax hint: if you use left and right before parentheses and other delimiters they autosize based on what is inside.
      $endgroup$
      – Ross Millikan
      Jan 9 at 5:43












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    2 Answers
    2






    active

    oldest

    votes








    2 Answers
    2






    active

    oldest

    votes









    active

    oldest

    votes






    active

    oldest

    votes









    2












    $begingroup$

    Sounds like you are looking for
    $$
    sum_{k = 0}^infty (2k+1) binom{2k}{k}left( frac18 right)^k,
    $$
    so consider differentiating
    $$
    f(x) = sum_{k = 0}^infty binom{2k}{k} x^{2k+1},
    $$
    and using $(1/8)^k = (1/sqrt8)^{2k}$






    share|cite|improve this answer









    $endgroup$


















      2












      $begingroup$

      Sounds like you are looking for
      $$
      sum_{k = 0}^infty (2k+1) binom{2k}{k}left( frac18 right)^k,
      $$
      so consider differentiating
      $$
      f(x) = sum_{k = 0}^infty binom{2k}{k} x^{2k+1},
      $$
      and using $(1/8)^k = (1/sqrt8)^{2k}$






      share|cite|improve this answer









      $endgroup$
















        2












        2








        2





        $begingroup$

        Sounds like you are looking for
        $$
        sum_{k = 0}^infty (2k+1) binom{2k}{k}left( frac18 right)^k,
        $$
        so consider differentiating
        $$
        f(x) = sum_{k = 0}^infty binom{2k}{k} x^{2k+1},
        $$
        and using $(1/8)^k = (1/sqrt8)^{2k}$






        share|cite|improve this answer









        $endgroup$



        Sounds like you are looking for
        $$
        sum_{k = 0}^infty (2k+1) binom{2k}{k}left( frac18 right)^k,
        $$
        so consider differentiating
        $$
        f(x) = sum_{k = 0}^infty binom{2k}{k} x^{2k+1},
        $$
        and using $(1/8)^k = (1/sqrt8)^{2k}$







        share|cite|improve this answer












        share|cite|improve this answer



        share|cite|improve this answer










        answered Jan 9 at 4:57









        gt6989bgt6989b

        35.2k22557




        35.2k22557























            0












            $begingroup$

            the series is the binomial expansion of
            $$
            bigg(1 - frac12 bigg)^{-frac32}
            $$



            which gives the sum $2sqrt{2}$






            share|cite|improve this answer











            $endgroup$









            • 1




              $begingroup$
              MathJax hint: if you use left and right before parentheses and other delimiters they autosize based on what is inside.
              $endgroup$
              – Ross Millikan
              Jan 9 at 5:43
















            0












            $begingroup$

            the series is the binomial expansion of
            $$
            bigg(1 - frac12 bigg)^{-frac32}
            $$



            which gives the sum $2sqrt{2}$






            share|cite|improve this answer











            $endgroup$









            • 1




              $begingroup$
              MathJax hint: if you use left and right before parentheses and other delimiters they autosize based on what is inside.
              $endgroup$
              – Ross Millikan
              Jan 9 at 5:43














            0












            0








            0





            $begingroup$

            the series is the binomial expansion of
            $$
            bigg(1 - frac12 bigg)^{-frac32}
            $$



            which gives the sum $2sqrt{2}$






            share|cite|improve this answer











            $endgroup$



            the series is the binomial expansion of
            $$
            bigg(1 - frac12 bigg)^{-frac32}
            $$



            which gives the sum $2sqrt{2}$







            share|cite|improve this answer














            share|cite|improve this answer



            share|cite|improve this answer








            edited Jan 9 at 5:42









            Ross Millikan

            301k24200375




            301k24200375










            answered Jan 9 at 5:27









            David HoldenDavid Holden

            14.9k21225




            14.9k21225








            • 1




              $begingroup$
              MathJax hint: if you use left and right before parentheses and other delimiters they autosize based on what is inside.
              $endgroup$
              – Ross Millikan
              Jan 9 at 5:43














            • 1




              $begingroup$
              MathJax hint: if you use left and right before parentheses and other delimiters they autosize based on what is inside.
              $endgroup$
              – Ross Millikan
              Jan 9 at 5:43








            1




            1




            $begingroup$
            MathJax hint: if you use left and right before parentheses and other delimiters they autosize based on what is inside.
            $endgroup$
            – Ross Millikan
            Jan 9 at 5:43




            $begingroup$
            MathJax hint: if you use left and right before parentheses and other delimiters they autosize based on what is inside.
            $endgroup$
            – Ross Millikan
            Jan 9 at 5:43


















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