series question 1/(n(n+2))?

Question

tell me how to solve the series of sum 1/(n(n+2)) don't just tell me answer, i really want to learn how to solve this type of problems. thx

Answer 1

The key to this is noting that:

1/(n(n+2)) = (1/2)[1/n - 1/(n+2)]

So summing (1/2)[1/n - 1/(n+2)] up from n = 1 to infinity we get:

(1/2)[(1/1 - 1/3)
+ (1/2 - 1/4)
+ (1/3 - 1/5)
+ ...]

All of the terms except for 1/1 and 1/2 are cancelled out, so the sum is (1/2)[1/1 + 1/2] = 3/4

Answer 2

write it as partial fractions as so

1/((n(n+2))=A/n+B/(n+2)

to find A and B mltiply by n(n+2) to get

1=A(n+2)+Bn

so A=1/2 and B=-1/2

Now you can write the original sum as the sum of two series

sum[1/(n(n+2))] = 1/2 sum[1/n] -1/2 sum[1/(n+2)]

If you write out these two series you will see that the third term of the second series is the same as the first term of the first, and that they cancel out.

This happens for all the rest of the terms too, leaving just the first two terms of the first series.

so your sum=1/2 *(1+1/2)=3/4

Answer 3

1/(n(n+2)=1/2[1/n-1/n+2]=1/2[1/n-1/(n+1)+1/(n+1)-1/(n+2)]so
sum fiom m to j of 1/(n(n+2)=1/2[1/m-1/(j+1)+1/(m+1)-1/(j+2)] so sum from 1 ti infinite is 1/2[1+1/2=3/4]

Answer 4

What the others said. You have to recognize the trick -- turning it from one series into two, with the two almost canceling each other out.