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Take for example the following: $${\left( {{1 \over 2}} \right)^n}$$ as n tends to infinity, the denominator tends to infinity, while the numerator stays the same (i.e. one). I thought $$\left( {{1 \over \infty }} \right) = 0$$ Yet Wikipedia says it equals one? Converging Sequence

Michael Lee
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    Your expression does tend to zero. It's the sum of all the terms along the way that converges to 1. – turkeyhundt Jan 14 '15 at 22:32
  • Oh, I see the difference now. But if one divided by infinity equals zero, then infinity multiplied by zero is one??? – Michael Lee Jan 14 '15 at 22:37
  • That's a little above my pay grade, but you can check this out. http://math.stackexchange.com/questions/28940/why-is-infinity-multiplied-by-zero-not-an-easy-zero-answer – turkeyhundt Jan 14 '15 at 22:39
  • @MichaelLee notice that multiplying the equation by $a\in\mathbb{R}$ you get that $0\cdot\infty=a$ but $a$ can be any real number hence limit of form $\infty\cdot 0$ can be any value. – kingW3 Jan 14 '15 at 22:44
  • @kingW3 I see and the numerator can be any other real number and infinity is not a real number. – Michael Lee Jan 15 '15 at 00:42
  • Regarding the title: the sequence tends to zero, the limit equals zero. – Hans Lundmark Jan 15 '15 at 07:29

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The summands $\frac1{2^n}$ tend to zero indeed. But the Wikipedia page you reference is about the series $\sum_{n=1}^\infty \frac1{2^n}$.

Hagen von Eitzen
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  • And of course, an infinite series can't possibly converge if the summand doesn't tend to zero! – Eff Jan 14 '15 at 22:37