Question

Evaluate \[\sum_{n=2}^{\infty}{\frac{1}{n(n+2)}}\]

Answer 1

1/n(n+2) = A/n + B/(n+2)
A(n+2) + Bn = 1
An + 2A +Bn = 1
A = 0.5 , B = -0.5

sum (1/2n - 1/2(n+2))
from n =2
now observe :

sum(1/2n) (from n =2) = 1/4 + 1/6 + 1/8 + 1/10 + 1/12 ...
sum(1/2(n+2)) (from n =2) = 1/8 + 1/10 + 1/12....

subtract the second from the first and get 1/4 + 1/6
= 5/12.

Answer 2

you haven't considered the last terms of both series.

Answer 3

what do you mean ?

i found a formula for this series too now :
http://upload.wikimedia.org/wikipedia/en/math/a/b/1/ab110a78934c413b000447358976b4ed.png
just subtract the first term (1/3) from this sum

Answer 4

hmm the integral converges... so I guess the sum also converges.

Answer 5

yes its 5/12 thanks