`sum_(n=1)^oo1/(n(n+1))`
`a_n=1/(n(n+1))`
Using partial fractions, we can write the n'th term as,
`a_n=1/n-1/(n+1)`
The n'th partial sum of the series `S_n` is,
`S_n=(1/1-1/(1+1))+(1/2-1/(2+1))+(1/3-1/(3+1))+............+(1/(n-1)-1/(n-1+1))+(1/n-1/(n+1))`
`S_n=(1-1/2)+(1/2-1/3)+(1/3-1/4)+............+(1/(n-1)-1/n)+(1/n-1/(n+1))`
This is telescoping form of the series,
`S_n=(1-1/(n+1))`
`sum_(n=1)^oo1/(n(n+1))=lim_(n->oo)S_n`
`=lim_(n->oo)(1-1/(n+1))`
`=1`
So the series converges.
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