A difficult sum?

$\sum _{ k=1 }^{ n }{ \frac { 1 }{ k(k+1) } } =\sum _{ k=1 }^{ n }{ \frac { k+1-1 }{ k(k+1) } } =\sum _{ n+1 }^{ n }{ \frac { 1 }{ k } -\frac { 1 }{ k+1 } }$

on replacing k by 1,2,3...,the equation condenses to $1-\frac { 1 }{ n+1 } =\frac { 99 }{ 100 }$

on solving, n=99

1/(k*(k+1)) = (1/k) - (1/k+1) ....... as such summation= 1 - (1/2) + (1/2) - (1/3) + (1/3) - (1/4) + (1/4) ............. - (1/(n+1)) = (99/100)...... which implies 1 - (1/(n+1)) = (99/100)..... simplifying we get n = 99 !!!!!!!!