find the sum of infinity 1+1/2(3/4)+1.3/2.4(3/4)^2
Answers
Answer:
By symmetry with the rest of the series, I have to suppose the first term is actually 1/(2·4) so the question is to find (we take a common 12 factor out of all terms)
12∑∞m=0∏mn=02n+12n+4
We can stick the inner part into Wolfram Alpha, just to see if it is reasonable to calculate, and we get that
∏mn=02n+12n+4=1π√Γ(m+32)Γ(m+3)
Once we know the answer, it is relatively easy to prove it, take out a common factor of 2 from each internal fraction, and multiply everything out:
∏mn=02(n+12)2(n+2)=∏mn=0n+12n+2=∏mn=0n+12(m+2)!
The product at the numerator is exactly Γ(m+32)Γ(12)=1π√Γ(m+32) , by the multiplicative property of Γ .
So now we just have to find (again, take out the common 1π√ factor):
12π√∑∞m=0Γ(m+32)Γ(m+3)
If you plug ∑∞m=0Γ(m+32)Γ(m+3) back to Alpha, the answer is π−−√ , but in this case I haven’t been able to find out the actual operations to get to this result. One possible way is to use duplication formula to get Γ(m+32)=π√Γ(2m+2)22n+1Γ(m+1) so the fraction becomes (as Γ(n+1)=n! ):
π−−√∑∞m=0(2m+1)!22m+1m!(m+2)!
In any case the raw result is
1 / 2
Step-by-step explanation:
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