Thursday, March 8, 2018

sequences and series - Calculating 1+frac13+frac1cdot33cdot6+frac1cdot3cdot53cdot6cdot9+frac1cdot3cdot5cdot73cdot6cdot9cdot12+dots?



How to find infinite sum How to find infinite sum 1+13+1336+135369+135736912+?



I can see that 3 cancels out after 1/3, but what next? I can't go further.


Answer



As the denominator of the nth term Tn is 36912(3n)=3nn!


(Setting the first term to be T0=1)



and the numerator of nth term is 135(2n1) which is a product of nth terms of an Arithmetic Series with common difference =2,


we can write 135(2n1)=12(121)(12n+1)(2n)


which suitably resembles the numerator of Generalized binomial coefficients


Tn=12(121)(12n+1)n!(23)n


So, here z=23,α=12 in (1+z)α


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