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anonymous
 4 years ago
Find the nonnegative number of c.
anonymous
 4 years ago
Find the nonnegative number of c.

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anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0\[f(n) = (1  \frac{1}{2})(1  \frac{1}{2^{2}})(1  \frac{1}{2^{3}}) ... ((1  \frac{1}{2^{n}})\] Find and prove a nonnegative number "c" such that f(n) > c for all n

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0c cannot be the largest c that will work, nor anywhere near it.

slaaibak
 4 years ago
Best ResponseYou've already chosen the best response.0What 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+%2811%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?

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0The question was given that P(n) is always greater than 0.288. I am not really sure either
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