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So in Euler's solution of the Basel problem he takes $\sin x= x\left(1-\frac{x^2}{\pi^2}\right)\left(1-\frac{x^2}{4\pi^2}\right)\cdots$

well, i was playing around with it and put $ x=\frac{\pi}{2}$

After manipulating quite a bit, $$\frac{2}{\pi}=\left(1-\frac{1}{2^2}\right)\left(1-\frac{1}{(2*2)^2}\right)\cdots\left(1-\frac{1}{(2n)^2}\right)\cdots$$ after simplifying a bit$$\frac{2}{\pi}=\left(\frac{3}{4}\right)\left(\frac{15}{16}\right)\left(\frac{35}{36}\right)\left(\frac{99}{100}\right)\cdots$$

I tried evaluating LHS by hand, and the partial products slowly do decrease towards RHS, and we know that they will not decrease forever because $$\lim_{n\to\infty} \frac{n^2-1}{n^2}=1$$ so can anyone help by writing a program which can do partial sum up to high values of n and tell me if its true. PS.I am in 10th standard do if im wrong anywhere please dont explain in too complicated mathematics. thanks

1 Answers1

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$\dfrac2{\pi} \approx 0.6366198$ is the correct limit of the product

As an illustration (not a proof) using R:

maxn  <- 2^15
plot(exp(cumsum(log(1 - (2*(1:maxn))^(-2)))), log="x")
abline(h=2/pi, col="red")

giving

enter image description here

Henry
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