Show that this sum equals (1/2)*Pi^(3/2)?
sum from k=0 to infinity
(5^(1+2*k)/((10349/77)^(1/2+k)) + 9^(1+2*k)/((16637/77)^(1/2+k)) + 45^(1+2*k)/(10033^(1/2+k))) * Γ(1/2+k) / ((1+2*k)*k!)
where Γ is the gamma function.
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- Scythian1950Lv 76 years agoFavorite Answer
Okay, let's plow through this a step at a time. Start with the series expansion of ArcSin(x):
∑ (k = 0 to ∞) (1 / 4^k (2k+1)) x^(2k+1) C(2k,k), ..which means ArcSin(√y) is:
∑ (k = 0 to ∞) (1 / 4^k (2k+1)) y^(k+1/2) C(2k,k),....which we could then use the Gamma identity:
Gamma(1/2 + k) =√π k! C(2k,k) / 4^k ... to come up with:
√π ArcSin[√y] = ∑ (k = 0 to ∞) (1 / (2k+1)k! ) y^(k+1/2) Gamma(1/2 + k)
which is starting to look familiar. Thus, the problem reduces to proving that:
ArcSin(a) + ArcSin(b) + ArcSin(c) = π/2... where
a = √ (25)(77)/(10349)
b = √ (81)(77)/(16637)
c = √ (2025)/(10033)
This is where this problem gets really annoying, because then it first has to be determined geometrically that given a, b, the formula for c in terms of a, b is:
√ ( 1 - a² - b² + 2a²b² - 2ab√((1 - a²)(1 - b²)) )
and so lo and behold, given a, b in this case, this formula does produce c. QED.
- Josh SwansonLv 66 years ago
@Scythian: for what it's worth, "This is where this problem gets really annoying, because then it first has to be determined geometrically that given a, b, the formula for c in terms of a, b is" isn't true. Pretty mindless algebraic manipulations work. Let A=ArcSin(a), etc., note Cos(A) = √(1-a^2), Sin(A) = a, etc. Taking Cos of both sides of your expression, it reduces to...
√((1-a^2)(1-b^2)(1-c^2)) - ab √(1-c^2) - bc √(1-a^2) - ca √(1-b^2)
Divide by the first term to get
1 =? ab/√((1-a^2)(1-b^2) + bc/√((1-b^2)(1-c^2)) + ca/√((1-c^2)(1-a^2))
It's tedious but easy to compute the terms on the RHS as, respectively,
which indeed sums to 1. In fact, we might be able to reverse engineer the problem this way. Let p=a/√(1-a^2), and likewise with q and r. We have pq = 77/208, qr = 81/208, rp = 50/208. You can check there is a unique (positive) triple (p, q, r) satisfying these conditions. You can use the quadratic equation to recover (a, b, c), from just picking three positive (say) rationals which sum to 1. Apply the ArcSin series and obfuscate the results a little to obtain something like the original expression.