#lim_(n->oo)(1/(1 xx 2)+1/(2 xx 3)+1/(3 xx4) + cdots + 1/(n(n+1)))#?

1 Answer
George C.
Sep 3, 2016

#lim_(N->oo) sum_(n=1)^N 1/(n(n+1)) = 1#

Explanation:

Note that:

#1/(n(n+1)) = ((n+1)-n)/(n(n+1)) = 1/n - 1/(n+1)#

So we find:

#sum_(n=1)^N 1/(n(n+1)) = sum_(n=1)^N (1/n - 1/(n+1))#

#color(white)(sum_(n=1)^N 1/(n(n+1))) = sum_(n=1)^N 1/n - sum_(n=1)^N 1/(n+1)#

#color(white)(sum_(n=1)^N 1/(n(n+1))) = sum_(n=1)^N 1/n - sum_(n=2)^(N+1) 1/n#

#color(white)(sum_(n=1)^N 1/(n(n+1))) = 1+color(red)(cancel(color(black)(sum_(n=2)^N 1/n))) - color(red)(cancel(color(black)(sum_(n=2)^N 1/n))) - 1/(N+1)#

#color(white)(sum_(n=1)^N 1/(n(n+1))) = 1 - 1/(N+1)#

So:

#sum_(n=1)^oo 1/(n(n+1)) = lim_(N->oo) sum_(n=1)^N 1/(n(n+1)) = lim_(N->oo) (1-1/(N+1)) = 1#

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