Q) Find the sum of the series 1/2 + 1/6 + 1/12 + 1/20 .....n terms where n tends to infinity.

A)
The series is:
1/1.2 + 1/2.3 + 1/3.4 ........

= summation from 1 to inf. of [1/(n)(n+1)]
Now, 1/(n)(n+1) = n+1-n / (n)(n+1)
= (1/n) - 1/(n+1)

summation will give:
1 + 1/2 + 1/3 + 1/4+ 1/n....... - 1/2 - 1/3 - 1/4 - 1/5....-1/n - 1/(n+1)
= 1 - 1/(n+1)
put n tends to infinity and you get
1 - 0
= 1

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