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1/1+1 2 +1 4 +1/1+2 2 +2 4 +1/1+3 2 +3 4 +......... infinite find 14s
The n th term in the series can be written as:S(n) = n/(1 + n² + )Our aim is to convert this into a form where we can cancel adjacent terms and finally reach at a numerical value.Therefore we should convert this into term1 - term2 form , by using partial fractions.S(n) = n/(1 + n² + ) = n/(S(n) = n/ (n² + 1)² - n² =By applying a² - b² = (a + b)(a - b)s(n) = n/ (n² + n + 1)(n² - n + 1)Now we can writen = 1/2 (n² + n + 1) - (n² - n + 1)Substituting this in S(n),S(n) = 0.5 x = 0.5 x - Therefore,Now adding each terms for n = 1 , 2 , 3 , ....Sum = 0.5 ( 1 - 1/3 + 1/3 - 1/7 + 1/7 - 1/13 ......)We can observe that the last partial fraction of nth term and first partial fraction of n+1 th term gets cancelled.This can be extended to all n as it goes till infinity.Therefore ,Sum = 0.5Therefore the sum of the series till infinity is 0.5
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