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Sum of a Telescoping Series (II)

n

6

The area of each colored region is

n

n+1

-

n-1

n

=

1

n(n+1)

, so

1

1·2

+

1

2·3

+⋯+

1

n(n+1)

=

n

n+1

and

∞

∑

n=1

1

n(n+1)

=1

.

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