How do i figure out this series?

Question

Examine the series:

The sum of: 1 over the expression [n(n+1)(n+2)].

How does this converge to 1/4? Thanks

Answer 1

1) make descomposition into partial fractions
a_n = 1/2*1/n -1(n+1) +1/2 *1/(n+2)
If you call
H_n = 1+1/2+1/3++++1/n

H_n= ln (n) + c + e_n where c is the so called Euler´s constant
(irrational number = approx0.57....) and e_n an expression with limit =0

So if we begin summing with n=1

the first term is = 1/2(ln n +c+e_n)
The second is = -(ln(n+1) +c +e_n+1 -1 ) as here the sum starts with the term 1/2
and the third is
1/2(ln(n+2)+c+e_n+2-1-1/2)
summing up

S_n = ln[ sqrt(n(n+2)]/(n+1)]+e´_n +1-1/2-1/4
e´_n is the sum of the three e and has limit 0 the ln also has limit 0 as [sqrt(n*(n+2)]/n has limit 1
So only remains S = 1-1/2-1/4 =1/4
I hope you can understand it