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`sum_(n=1)^oo 4/(n(n+2))` Find the sum of the convergent series.

`sum_(n=1)^oo4/(n(n+2))`


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


`a_n=4(1/(2n)-1/(2(n+2)))`  


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


Now the n'th partial sum is,


`S_n=(2/1-2/(1+2))+(2/2-2/(2+2))+(2/3-2/(3+2))+(2/4-2/(4+2))+(2/5-2/(5+2))+............`


`S_n=(2-2/3)+(1-1/2)+(2/3-2/5)+(1/2-1/3)+(2/5-2/7)+......`


`S_n=(2+1)`  all other terms cancel out


`S_n=3`


`sum_(n=1)^oo4/(n(n+2))=3`

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