# Estimate the infinite product

Explain: What does the cartesian symbols mean? Whats going on in here? What topic of calculus is this?

$\large\prod_{i = 1}^\infty \left(1 + \frac{1}i \right)^{(-1)^{-i}} = \ \lim_{n \to\infty} \prod_{i = 1}^n \left(1 + \frac{1}i \right)^{(-1)^{-i}}$

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Was this copy/paste from an online exam? The verbiage looks like it... – apnorton Feb 3 '13 at 23:30
What do you mean by "the cartesian symbols"? You seem to know that $\prod$ signifies a product. – Gerry Myerson Feb 3 '13 at 23:31
It looks like the upper limit on the right should be $n$, not $\infty$ and this is just a definition of what the left side means. – Ross Millikan Feb 3 '13 at 23:33

$$\prod_{i=1}^{N} \left(1 + \dfrac1i\right)^{(-1)^i} = \left(\dfrac21\right)^{-1}\left(\dfrac32\right)^{1}\left(\dfrac43\right)^{-1}\left(\dfrac54\right)^{1} \cdots \left(\dfrac{N}{N-1}\right)^{-1^{N-1}}\left(\dfrac{N+1}N\right)^{(-1)^N}$$ Hence, $$\prod_{i=1}^{2N} \left(1 + \dfrac1i\right)^{(-1)^i} = \dfrac12 \dfrac32 \dfrac34 \dfrac54 \dfrac56 \dfrac76 \cdots \dfrac{2N-1}{2N}\dfrac{2N+1}{2N}$$ Recall that $$\int_0^{\pi/2} \sin^{2N}(x) dx = \dfrac{2N-1}{2N}\dfrac{2N-3}{2N-2} \cdots \dfrac34 \dfrac12 \dfrac{\pi}2$$ and $$\int_0^{\pi/2} \sin^{2N+1}(x) dx = \dfrac{2N}{2N+1}\dfrac{2N-2}{2N-1}\dfrac{2N-4}{2N-3} \cdots \dfrac45 \dfrac23$$ Note that $$\int_0^{\pi/2} \sin^{2N-1}(x) dx > \int_0^{\pi/2} \sin^{2N}(x) dx > \int_0^{\pi/2} \sin^{2N+1}(x) dx$$ Hence, $$\dfrac{2N-2}{2N-1}\dfrac{2N-4}{2N-3} \cdots \dfrac45 \dfrac23 > \dfrac{2N-1}{2N}\dfrac{2N-3}{2N-2} \cdots \dfrac34 \dfrac12 \dfrac{\pi}2 > \dfrac{2N}{2N+1}\dfrac{2N-2}{2N-1} \cdots \dfrac45 \dfrac23$$ $$\dfrac{2N}{2N-1}\dfrac{2N-2}{2N-1}\dfrac{2N-2}{2N-3}\dfrac{2N-4}{2N-3} \cdots \dfrac45 \dfrac43 \dfrac23 \dfrac21 > \dfrac{\pi}2 > \dfrac{2N}{2N+1} \dfrac{2N}{2N-1}\dfrac{2N-2}{2N-1}\dfrac{2N-2}{2N-3}\dfrac{2N-4}{2N-3} \cdots \dfrac45 \dfrac43 \dfrac23 \dfrac21$$ Can you now conclude what the answer should be?