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So I have to find the limit of $\frac 1 {1\times3} +\frac 1 {3\times5} + .. + \frac 1 {(2n-1)\times (2n+1)}$ as $n$ approaches infinity ... I don't know how to express the $\frac 1{1\times3} +\frac 1 {3\times5}$ part ?

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HINT $$\dfrac1{(2n-1)(2n+1)} = \dfrac12 \left(\dfrac1{2n-1} - \dfrac1{2n+1} \right)$$ And use telescopic summation/cancellation.

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the answer is zero then..but why,in my book it is 1/2 :/ –  ewwe Feb 3 '13 at 10:33
    
Not $0$. Almost everything cancels. But look at the "left part" of the first term. –  André Nicolas Feb 3 '13 at 11:11

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