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I was given the following series $\ S(n) = 1+\frac{1}{2}\cdot \frac{1}{2}+ \frac{3*1}{4*2}\cdot\left ( \frac{1}{2}\right )^2 + \frac{5*3*1}{6*4*2}\left ( \frac{1}{2} \right )^3 + ... $

i noticed that each individual term looks very similar to the solution for a wallis integral

Thus $$ S(n) = \frac{2}{\pi}\int_{0}^{\pi/2}\frac{sin^0x}{2^0} + \frac{sin^2x}{2^1}+\frac{sin^4x}{2^2}+\frac{sin^6x}{2^3}+...dx $$ Which appeared to be a geometric series, Hence $$ S(n) = \frac{2}{\pi}\int_{0}^{\pi/2} \frac{(\frac{sin^2x}{2})^n-1}{\frac{sin^2x}{2}-1} dx $$ Which after some algebraic simplifications gives me this horrible integral at which i am stuck $$ = \frac{2^{2 - n}}{\pi}\int_{0}^{\pi/2}\frac{sin^{2n}x - 2^n }{sin^2x - 2} dx $$

So how can this integral be solved ? is it non-elementary integral ? Or is there another approach to solve this problem ?

Edit: Thanks for your response, i thought at first that the question wanted me to find the sum of the series to the n-th term,

In continuing to my previous approach i would be using infinite geometric series formula instead

$$ S =\frac{2}{π} \int_{0}^{\pi/2} \frac{dx}{1 - \frac{sin^2x}{2}} = √2 $$ Which can be easily solved by different techniques !

Amr Elmowaled
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  • Have you tried change of variable $t = \sin x$?. The integrand transfrom into $\frac{t^n-2^n}{t-2}$ which it is equal to $t^{n-1} + t^{n-2}\cdot 2 + t^{n-3} \cdot 2^2 +...+2^{n-1}$. I didn't check if this work, but could be a good strategy. – Senna Mar 12 '22 at 11:52
  • See https://math.stackexchange.com/questions/746388/calculating-1-frac13-frac1-cdot33-cdot6-frac1-cdot3-cdot53-cdot6-cdot – lab bhattacharjee Mar 12 '22 at 12:00

1 Answers1

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Examine the coefficients more closely:

$$\frac{5\cdot3\cdot1}{6\cdot4\cdot2}=\frac{6!}{(6\cdot4\cdot2)^2}=\frac{6!}{(2^3\cdot 3!)^2}=\frac{6!}{(3!)(3!)}\cdot\frac{1}{2^{2\cdot 3}}=\binom{6}{3}\frac{1}{2^{2\cdot3}}$$

If the pattern is consistent as you suggest, we have the $n$th coefficient (considered as a power series in $1/2$):

$$\binom{2n}{n}\frac{1}{2^{2n}}$$

Which you may recognise (from the derivation of the $\arcsin$ series) as the binomial series of the square root function: $$(1-x)^{-1/2}$$

So, this is just $(1-1/2)^{-1/2}=\sqrt{2}$.

N.B: Why is that the series of $(1-x)^{-1/2}$? The derivatives of this function $f(x)=(1-x)^{-1/2}$ at $x=0$ are of the form (don't forget the chain rule with the $-x$ in the middle!):

$$\begin{align}f(0)&=1^{-1/2}=1\\f'(0)&=-\frac{1}{2}(-1)\cdot1^{-3/2}=\frac{1}{2}\\f''(0)&=-\frac{1}{2}(-1)\cdot-\frac{3}{2}(-1)\cdot1^{-5/2}=\frac{1\cdot3}{2\cdot2}\\&\cdots\\f^{(n)}(x)&=\frac{(2n-1)!!}{2^n}(1-x)^{-(2n+1)/2},\,f^{(n)}(0)=\frac{(2n-1)!!}{2^n}\end{align}$$

Try showing this by induction, and then using $f(x)=\sum_{n=0}^\infty\frac{f^{(n)}(0)}{n!}\cdot x^n$ try showing the coefficients match the form I gave before.

FShrike
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