hi Let P = [(2^3-1)/(2^3+1)][(3^3-1)/(3^3+1)]........[(n^3-1)/(n^3+1)] n=2,3,4,...... Find Limit P when N tends to infinity.
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Originally Posted by champrock hi Let P = [(2^3-1)/(2^3+1)][(3^3-1)/(3^3+1)]........[(n^3-1)/(n^3+1)] n=2,3,4,...... Find Limit P when N tends to infinity. Factorise as , and notice that . You will then find that you have a telescoping product.
what should i do with n^2 - n + 1 ? I was able to cancel the (n-1)/(n+1) terms. but dont know what to do with n^2 - n + 1 and n^2 + n + 1
You easily undestand if you write the sequence of terms... ... and it is evident you can simplify both in numerator and denominator [3 with 3, 4 with 4, ... , 7 with 7, 13 with 13, 21 with 21,...] Finally is... Very nice!... Kind regards
Originally Posted by champrock what should i do with n^2 - n + 1 ? I was able to cancel the (n-1)/(n+1) terms. but dont know what to do with n^2 - n + 1 and n^2 + n + 1 Use the hint I gave before: . That tells you that the term in the numerator of each fraction cancels with the term in the denominator of the next one.
I think Only 1/3 is left. All the other terms cancel out. How are you getting 2/3
The term of order n is... For is... Kind regards
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