Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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

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

share|improve this question
    
Was this copy/paste from an online exam? The verbiage looks like it... –  anorton 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
2  
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

1 Answer 1

up vote 1 down vote accepted

$$\prod_{i=1}^{N} \left(1 + \dfrac1i\right)^{(-1)^i} = \left(\dfrac21\right)^{-1}\left(\dfrac32\right)^{1}\left(\dfrac43\right)^{-1}\left(\dfrac54\right)^{1} \cdots \left(\dfrac{N}{N-1}\right)^{-1^{N-1}}\left(\dfrac{N+1}N\right)^{(-1)^N}$$ 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?

share|improve this answer

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.