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Old 2003-04-29, 19:47   #3
S80780
 
Jan 2003
far from M40

7D16 Posts
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Yes, it is.

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

So, the kth partial - sum has the value

1 - 1/(k+1)

which's limit for k to infinity is 1 q.e.d.

Sums like these, where a summand is (partly) nihilized by its successor are called telescope sums.

Benjamin
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