1my intuition says d answer is close to 0...coz d answer fluctates between positive n negativ...isnt it,sir??
can i do by breaking d denominator 1's into (a+b)(a-b)..????
Find the product of these numbers...
(1-\frac{1}{2^2})(1-\frac{1}{3^2})(1-\frac{1}{5^2})(1-\frac{1}{7^2})(1-\frac{1}{11^2})... till infinity...
Given that \frac{1}{1^2}+\frac{1}{2^2}+\frac{1}{3^2}+\frac{1}{4^2}+\frac{1}{5^2}+\frac{1}{6^2}...=\pi^2/6
May seem very difficult.. but think closely.. it is not [1]
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3 Answers
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62btw for the origianl question, your analysis is corrct..
denominoator would have become infinitely large.. hence the number would be zero.. :)