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\[\lim_{n \rightarrow \infty}(\frac{ 1 }{ 1*3 }+\frac{ 1 }{ 3*5 }+......\frac{ 1 }{ (2n-1)(2n+1) })\]

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2=3-1 so, 1/1*3 = 1/2(2/1*3) = 1/2((3-1)/3*1) = 1/2[1-1/3] do this for every term. should i do it in latex, or you got it ?
\[\frac{1}{1 \times 3}=\frac{1}{2} \times \frac{2}{1 \times 3}=\frac{1}{2} \times[\frac{3-1}{1 \times 3}]=\frac{1}{2} \times[\frac{1}{1 }-\frac{1}{ 3}]\] do this for every term.

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Other answers:

2nd term = 1/3 -1/5 notice 1/3 will get cancelled, and if you go on, all the terms excepts 1st and last will get cancelled.
Would u Call This Partial Fraction Decomposition?
yes. i would.
it would be n /(2n+1) ryt? @hartnn
then limit=... ?
use the telescopic's principle, it can be 1/2 (1 - 1/(2n+1)) 1/2 (2n/(2n+1)) just look the cofficient of n, they are same) :)

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