Tuesday, June 21, 2016

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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