anonymous
  • anonymous
Find the non-negative number of c.
Meta-math
jamiebookeater
  • jamiebookeater
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anonymous
  • anonymous
\[f(n) = (1 - \frac{1}{2})(1 - \frac{1}{2^{2}})(1 - \frac{1}{2^{3}}) ... ((1 - \frac{1}{2^{n}})\] Find and prove a non-negative number "c" such that f(n) > c for all n
anonymous
  • anonymous
c cannot be the largest c that will work, nor anywhere near it.
slaaibak
  • slaaibak
What do you mean by c cannot be the largest c that will work, nor anywhere near it ? If we compute the product like this: http://www.wolframalpha.com/input/?i=product+from+1+to+infinity+of+%281-1%2F2^%28n%29%29 we see it "converges" to: 0.288788095086602421278899721929 so if we must choose a c value, it can be anywhere between 0 and 0.288788095086602421278899721929 Or am I misreading things?

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anonymous
  • anonymous
The question was given that P(n) is always greater than 0.288. I am not really sure either

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