Infinite sum/difference of ratios of products

Probability Level 4

1 − 3 4 + 3 × 5 4 × 8 − 3 × 5 × 7 4 × 8 × 12 + 3 × 5 × 7 × 9 4 × 8 × 12 × 16 − ⋯ ? 1 - \frac{ 3 } { 4 } + \frac{3 \times 5 } { 4 \times 8 } - \frac{ 3 \times 5 \times 7 } { 4 \times 8 \times 12 } + \frac{ 3 \times 5 \times 7 \times 9 } { 4 \times 8 \times 12 \times 16 } - \cdots \, ?

What is the exact value of the expression above?

Give your answer to 3 decimal places.


The answer is 0.544331053951817356306719375425018370151519775390625.

Calvin Lin by Calvin Lin

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1 solution

Calvin Lin Staff
May 13, 2014

Observe that each coefficient is of the form ( 1 2 ) n ( − 3 2 n ) \left(\frac{ 1}{2} \right)^n { -\frac{3}{2} \choose n } . As such, this encourages us to consider the binomial expansion of ( 1 + x ) − 3 2 ( 1 + x) ^ {-\frac{3}{2}} , which is

1 − 3 x 2 + 15 x 2 8 − 35 x 3 16 + … . 1-\frac{ 3 x}{2}+\frac{15 x^2}{8}-\frac{35 x^3}{16} + \ldots .

Substituting x = 1 2 x = \frac{1}{2} , we get that the value of the expression is 3 2 − 3 2 = 8 27 \frac{ 3}{2} ^ { - \frac{3}{2} } = \sqrt{ \frac{8}{27} } .

12 Helpful 0 Interesting 0 Brilliant 0 Confused

I didn't immediately notice this approach. To be honest, I use the 'wrong' approach but it turned out to the correct result. I noticed that the expression in term of sum of series was ∑ n = 0 ∞ ( − 1 ) n ( 2 n + 1 ) ! ! 2 2 n n ! \sum_{n=0}^\infty (-1)^n\frac{(2n+1)!!}{2^{2n} n!} After some manipulations using relation of double factorial and gamma function, then formed the series into beta function and used geometric series, I got this integral − 2 π ∫ 0 1 x 1 / 2 ( 1 − x ) 3 / 2 ( 2 + x ) d x -\frac{2}{\pi}\int_0^1\frac{x^{1/2}}{(1-x)^{3/2}(2+x)}\,dx then I used this formula to evaluate the integral ∫ 0 1 x q − 1 ( 1 − x ) q ( 1 + p x ) d x = π ( 1 + p ) q sin ⁡ q π \int_0^1\frac{x^{q-1}}{(1-x)^{q}(1+px)}\,dx=\frac{\pi}{(1+p)^q\sin q\pi} Putting q = 3 2 q=\frac{3}{2} and p = 1 2 p=\frac{1}{2} , I got ∑ n = 0 ∞ ( − 1 ) n ( 2 n + 1 ) ! ! 2 2 n n ! = 8 27 \sum_{n=0}^\infty (-1)^n\frac{(2n+1)!!}{2^{2n} n!}=\sqrt{\frac{8}{27}} The problem is the formula only holds for 0 < q < 1 0<q<1 and p > − 1 p>-1 . I don't get it. Plucking the integral to Mathematica, it returns the output: "Integral does not converge".

Anastasiya Romanova - 6 years, 7 months ago

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Sorry, I spot my mistake.

Anastasiya Romanova - 6 years, 7 months ago

thanks we got it

Shiva Sai - 6 years, 10 months ago

Is it necessary to show that it converges in the first place?

Joel Tan - 6 years, 7 months ago

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