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Mathematics 7 Online
OpenStudy (anonymous):

What is the closed form of:\[\sum_{n=1}^{k} \frac{1}{n(n+1)}\]

OpenStudy (bahrom7893):

looks sooo familiar lol

OpenStudy (anonymous):

well no recursive cancellation like before XD

OpenStudy (anonymous):

1-{1/(n+1)} correct?

OpenStudy (anonymous):

closed for so that has to be in variable k

OpenStudy (anonymous):

it not infinite series

OpenStudy (anonymous):

1-{1/(k+1)}

OpenStudy (bahrom7893):

doncatch can u explain how u got that..

OpenStudy (anonymous):

doncatch is right

OpenStudy (anonymous):

it is easy ..... 1/n.(n+1) = 1/n - 1/(n+1) so sum = (1/2-1/3)+(1/3-1/4)+(1/4-1/5)+....+(1/k-1/k+1) so sum = 1-1/(1+K) Di u unerstan?

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