How to find infinite sum How to find infinite sum 1+13+1⋅33⋅6+1⋅3⋅53⋅6⋅9+1⋅3⋅5⋅73⋅6⋅9⋅12+…?
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 3⋅6⋅9⋅12⋯(3n)=3n⋅n!
(Setting the first term to be T0=1)
and the numerator of nth term is 1⋅3⋅5⋯(2n−1) which is a product of nth terms of an Arithmetic Series with common difference =2,
we can write
1⋅3⋅5⋯(2n−1)=−12⋅(−12−1)⋯(−12−n+1)⋅(−2n)
which suitably resembles the numerator of Generalized binomial coefficients
⟹Tn=−12⋅(−12−1)⋯(−12−n+1)n!(−23)n
So, here z=−23,α=−12 in (1+z)α
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