11/18 (0.61111…)

Each term 1/(n(n+3)) can be rewritten as two separate terms:

A/n + B/(n+3)

To find the values of A and B we force them back into a single fraction and make sure it’s equal to the original fraction:

(A(n+3) + Bn)/ (n(n+3)) = 1/(n(n+3))

(A+B)n + 3A = 1

The n term must disappear as there isn’t one on the right-hand side, so A+B=0.

3A=1, so A=1/3, and so B=-1/3

So each term 1/(n(n+3)) is equivalent to 1/3n - 1/(3(n+3)). For example the first term 1/4 is equal to (1/3 - 1/12), the second term 1/10 is equal to (1/6 - 1/15), etc.

This might seem to be unnecessarily complicating the infinite sum, however you might notice that the negative part of each term will appear again as the positive part, three terms later:

(1/3 - 1/12) + (1/6 - 1/15) + (1/9 - 1/18) + (1/12 - 1/21) + (1/15 - 1/24) + (1/18 - 1/27) …

Virtually everything will cancel out. Only the positive parts of the first three terms will remain, and can easily be added together to find the final answer:

1/3 + 1/6 + 1/9 = 11/18, or 0.6111…

If you were to add up successive terms on a spreadsheet you would find that it does indeed approach this sum, but it does so very very slowly. It only reaches 0.6 after about 90 terms, 0.61 after about 900 terms, 0.611 after about 9000 terms, 0.6111 after about 90000 terms etc.