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What is the closed form of:\[\sum_{n=1}^{k} \frac{1}{n(n+1)}\]
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looks sooo familiar lol
well no recursive cancellation like before XD
1-{1/(n+1)} correct?
closed for so that has to be in variable k
it not infinite series
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1-{1/(k+1)}
doncatch can u explain how u got that..
doncatch is right
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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