`sum_(n=1)^oo 1/(n(n+2))` Verify that the infinite series converges

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

We can write down the n'th term of the series as,

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

Using partial fractions,

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

The n'th partial sum of the series is:

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

`S_n=(1/2-1/6)+(1/4-1/8)+(1/6-1/10)+..........+(1/(2(n-1))-1/(2(n+1)))+(1/(2n)-1/(2(n+2)))`

This is telescoping form of the series,

`S_n=(1/2+1/4-1/(2(n+1))-1/(2(n+2)))`

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

`=lim_(n->oo)(1/2+1/4-1/(2(n+1))-1/(2(n+2)))`

`=(1/2+1/4)`

`=3/4`

So the series converges.

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