You need to notice that general term of sum is of form `1/(n*(n+1)), ` hence, you should use partial fraction decomposition such that:

`1/(n*(n+1)) = A/n + B/(n+1)`

Bringing all terms to a common denominator yields:

`1 = A(n+1) + Bn`

`1 = An + A+ Bn`

Collecting like terms yields:

`1 = n(A+B) + A`

Equating coefficients of like powers both sides yields:

`A+B = 0 =gt A=-B`

`A=1 =gt B=-1`

Substituting 1 for A and -1 for B yields:

`1/(n*(n+1)) = 1/n- 1/(n+1)`

Hence, you may write each term of sum such that:

`1/(1*2) = 1/1 - 1/2`

`1/(2*3) = 1/2 - 1/3`

`1/(3*4) = 1/3 - 1/4`

............................

`1/(n*(n+1)) = 1/n - 1/(n+1)`

Adding these terms yields:

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

Reducing like terms yields:

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

You need to evaluate the limit of the sum such that:

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

`lim_(n-gtoo) (1 - 1/(n+1)) =lim_(n-gtoo) 1 -lim_(n-gtoo)1/(n+1)`

`lim_(n-gtoo) (1 - 1/(n+1)) = 1 - 1/oo`

`lim_(n-gtoo) (1 - 1/(n+1)) = 1 -` 0

`lim_(n-gtoo) (1 - 1/(n+1)) = 1`

**Hence, evaluating the limit of the sum yields `lim_(n-gtoo) (1/(1*2) + 1/(2*3) +...+1/(n*(n+1))) = 1.` **

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