How to find infinite sum How to find infinite sum 1+13+1⋅33⋅6+1⋅4⋅53⋅6⋅9+1⋅3⋅5⋅73⋅6⋅9⋅12+…

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How to find infinite sum How to find infinite sum  1+13+1⋅33⋅6+1⋅4⋅53⋅6⋅9+1⋅3⋅5⋅73⋅6⋅9⋅12+…

Suhadolahbb · Accepted 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

Daniel Cormack · Accepted Answer

Using Generalized Binomial Expansion,  (1+x)n=1+nx+n(n−1)2!x2+n(n−1)(n−2)3!x3+…  given the converge holds  Comparing with given Series nx=13⇒n2x2=… (1)  and n(n−1)2!x2=1⋅33⋅6 (2)  Divide (2) by (1) to find n=−12 and consequently x=−23  Observe that these values satisfy the next two terms, too.  Hence, the sum follows

alenahelenash · Accepted Answer

Consider denominator and numerator separately at first, Gn=2n∏m=1nn−1/2, Fn=13nn! Thus we have Tn=∏m=1nm−1/2n!(23)n or T=∑n=0∞∏m=1nm−1/2n!(23)n The final form of the series is ∑n=0∞Γ(n+1/2)πn!(23)n=3

How to find infinite sum How to find infinite sum 1+\frac{1}{3}+\frac{1\cdot3}{3\cdot6}+\frac{1\cdot4\cdot5}{3\cdot6\cdot9}+\frac{1\cdot3\cdot5\cdot7}{3\cdot6\cdot9\cdot12}+...

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