How do you find the nth partial sum, determine whether the series converges and find the sum when it exists given #1/(1*2)+1/(2*3)+...+1/(n(n+1))+...#?

2 Answers
Andrea S.
Apr 4, 2018

#sum_(n=1)^oo 1/(n(n+1)) = 1#

Explanation:

Given the series:

#sum_(n=1)^oo 1/(n(n+1))#

we can determine that is convergent using the direct comparison test since:

#1/(n(n+1)) < 1/n^2#

and the series:

#sum_(n=1)^oo 1/n^2#

is convergent based on the #p#-series test.

Decompose now the general term using partial fractions:

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

and note that in the partial sum:

#s_n = (-1/(n+1) +1/n) + (-1/n +1/(n-1)) + ... + (-1/3+1/2) + (-1/2 +1)#

all terms cancel each other except the first and the last, so:

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

and then:

#lim_(n->oo) s_n = 1#

is the sum of the series.

Narad T.
Apr 4, 2018

The answer is #=1#

Explanation:

#sum_(k=1)^n1/(k(k+1))=sum_(k=1)^n1/k-sum_(k=1)^n1/(k+1)#

#=sum_(k=1)^n1/k-sum_(k=2)^(n+1)1/(k)#

#=1+cancel(sum_(k=2)^n1/k)-cancel(sum_(k=2)^n1/k)-1/(n+1)#

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

#=n/(n+1)#

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

As

#lim_(n->+oo)1/n=0#

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