The series is of the form
#sum_(k=1)^n2/((k+1)(k+2))#
Perform the partial fraction decomposition
#2/((k+1)(k+2))=A/(k+1)+B/(k+2)#
#=(A(k+2)+B(k+1))/((k+1)(k+2))#
The denominators are the same, compare the numerators
#2=(A(k+2)+B(k+1))#
Let #k=-1#, #=>#, #2=A#
Let #k=-2#, #=>#, #2=-B#
Therefore,
#2/((k+1)(k+2))=2/(k+1)-2/(k+2)#
So, this is a telescoping series
#sum_(k=1)^n2/((k+1)(k+2))=sum_(k=1)^n2/(k+1)-sum_(k=1)^n2/(k+2)#
When #k=1#, #=>#, #u_1=2/2-cancel(2/3)#
When #k=2#, #=>#, #u_1=cancel(2/3)-cancel(2/4)#
When #k=3#, #=>#, #u_1=cancel(2/4)-cancel(2/5)#
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.
.
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When #k=n-1#, #=>#, #u_(n-1)=cancel(2/n)-cancel(2/(n+1))#
When #k=n#, #=>#, #u_(n)=cancel(2/(n+1))-2/(n+2)#
Adding all the terms
#sum_(k=1)^n2/((k+1)(k+2))=2/2-2/(n+2)=n/(n+2)#
And the limit is
#lim_(n->+oo)sum_(k=1)^n2/((k+1)(k+2))=lim_(n->+oo)n/(n+2)#
#=lim_(n->+oo)1/(1+2/n)#
#=1#
As
#lim_(n->+oo)(2/n)=0#
