A series is given as follows 16+56⋅12+5⋅86⋅12⋅18+5⋅8⋅116⋅12⋅18⋅24+…

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A series is given as follows  16+56⋅12+5⋅86⋅12⋅18+5⋅8⋅116⋅12⋅18⋅24+…

rakije2v · Accepted Answer

∑n=1∞12∏i=1n3i−16i=12∑n=1∞12nn!∏i=1n3i−13  =12∑n=1∞1(−2)nn!∏i=1n1−3i3  =12∑n=1∞1(−2)nn!∏i=1n(13−i)  =[12∑n=1∞xnn!∏i=1n(13−i)]x=−12  Now note that for n≥1, ∏i=1n(13−i)=f(n)(0), where f(x)=(x+1)−23. We can manipulate the expression a little further to see it as a Taylor series for f.  ∑n=1∞12∏i=1n3i−16i=[12∑n=0∞xnn!∏i=1n(13−i)−12]x=−12  =[12∑n=0∞xnn!f(n)(0)]x=−12−12  =12f(−12)

RizerMix · Accepted Answer

Consider the binomial expansion of (1+x)n=1+nx+n(n−1)2x2+... Multiplying the series term by term by 2 On comparing, nx=13 and n(n−1)2x2=53.12 Solving for n,x we get n=−23, x=−12 So the sum becomes [(12)−23−1]=223−1=413−1 since 2 has been multiplied before divide the result by 2

nick1337 · Accepted Answer

For any n∈N+ we have ∏k=1n(3k+2)=3nΓ(n+53)Γ(53) and the value of the RHS at n=0 equals 1. It follows that the whole series can be represented as S=∑n≥016n+1(n+1)!⋅3nΓ(n+53)Γ(53) =∑n≥016⋅2n⋅Γ(n+53)Γ(53)Γ(n+2) or, by using Euler&#039;s beta function and the reflection formula for the Γ function, as ∑n≥016Γ(13)Γ(53)⋅B(n+53,13)2n=38π∑n≥0∫01xn+2/3(1−x)−2/32ndx from which: S=34π∫01x2/3(1−x)2/3(2−x)dx=34π∫0+∞z2/3(1+z)(2+z) and: S=34π(−∫0+∞z2/3z(1+z)dz+2∫0+∞z2/3z(2+z)dz) Euler&#039;s beta function then leads to: S=123−12≈0.2937

A series is given as follows \frac{1}{6}+\frac{5}{6\cdot12}+\frac{5\cdot8}{6\cdot12\cdot18}+\frac{5\cdot8\cdot11}{6\cdot12\cdot18\cdot24}+...

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