`sum_(n=1)^oo 1/((2n+1)(2n+3))` Find the sum of the convergent series.

`sum_(n=1)^oo1/((2n+1)(2n+3))`

Using partial fractions, we can write the n'th term of the sequence as,

`a_n=1/(2(2n+1))-1/(2(2n+3))`

Now the n'th partial sum is,

`S_n=(1/(2(2+1))-1/(2(2+3)))+(1/(2(2*2+1))-1/(2(2*2+3)))+(1/(2(2*3+1))-1/(2(2*3+3)))+.............+(1/(2(2n+1))-1/(2(2n+3)))`

`S_n=(1/6-1/10)+(1/10-1/14)+(1/14-1/18)+........+(1/(2(2n+1))-1/(2(2n+3)))`

`S_n=(1/6-1/(2(2n+3)))`

`sum_(n=1)^oo1/((2n+1)(2n+3))=lim_(n->oo)S_n`

`=lim_(n->oo)(1/6-1/(2(2n+3)))`

`=1/6`

Comments

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