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

Duncan Reed

Duncan Reed

Answered

2022-01-21

A series is given as follows
16+5612+5861218+58116121824+

Answer & Explanation

rakije2v

rakije2v

Expert

2022-01-22Added 12 answers

n=112i=1n3i16i=12n=112nn!i=1n3i13
=12n=11(2)nn!i=1n13i3
=12n=11(2)nn!i=1n(13i)
=[12n=1xnn!i=1n(13i)]x=12
Now note that for n1, i=1n(13i)=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=112i=1n3i16i=[12n=0xnn!i=1n(13i)12]x=12
=[12n=0xnn!f(n)(0)]x=1212
=12f(12)
RizerMix

RizerMix

Expert

2022-01-27Added 437 answers

Consider the binomial expansion of (1+x)n=1+nx+n(n1)2x2+... Multiplying the series term by term by 2 On comparing, nx=13 and n(n1)2x2=53.12 Solving for n,x we get n=23, x=12 So the sum becomes [(12)231]=2231=4131 since 2 has been multiplied before divide the result by 2
nick1337

nick1337

Expert

2022-01-27Added 573 answers

For any nN+ 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=n016n+1(n+1)!3nΓ(n+53)Γ(53)
=n0162nΓ(n+53)Γ(53)Γ(n+2)
or, by using Euler's beta function and the reflection formula for the Γ function, as
n016Γ(13)Γ(53)B(n+53,13)2n=38πn001xn+2/3(1x)2/32ndx
from which:
S=34π01x2/3(1x)2/3(2x)dx=34π0+z2/3(1+z)(2+z)
and:
S=34π(0+z2/3z(1+z)dz+20+z2/3z(2+z)dz)
Euler's beta function then leads to:
S=123120.2937

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