Answer: The partial fraction decomposition is \(\frac{4}{r(r+1)(r+2)}=\frac{2}{r}-\frac{4}{r+1}+\frac{2}{r+2}\).

Hence

\(\sum_{r=1}^{n} \frac{4}{r(r+1)(r+2)}=1-\frac{2}{n+1}+\frac{2}{n+2}\).

So

\(\sum_{r=1}^{\infty} \frac{4}{r(r+1)(r+2)}=1\).

Let

\(\frac{4}{r(r+1)(r+2)}=\frac{A}{r}+\frac{B}{r+1}+\frac{C}{r+2}\).

Multiplying through by \(r(r+1)(r+2)\) gives

\(4=A(r+1)(r+2)+Br(r+2)+Cr(r+1).\)

Expanding:

\(4=A(r^2+3r+2)+B(r^2+2r)+C(r^2+r).\)

Collecting coefficients of powers of \(r\):

\(4=(A+B+C)r^2+(3A+2B+C)r+2A.\)

So

• \(A+B+C=0\)
• \(3A+2B+C=0\)
• \(2A=4\)

Hence \(A=2\). Then \(2+B+C=0\), so \(B+C=-2\), and \(6+2B+C=0\), so \(2B+C=-6\). Subtracting gives \(B=-4\), and therefore \(C=2\).

Thus

\(\frac{4}{r(r+1)(r+2)}=\frac{2}{r}-\frac{4}{r+1}+\frac{2}{r+2}.\)

Now sum from \(r=1\) to \(n\):

\(\sum_{r=1}^{n}\left(\frac{2}{r}-\frac{4}{r+1}+\frac{2}{r+2}\right).\)

Writing out the first few terms shows the cancellation:

\((2-2+\tfrac{2}{3})+(1-\tfrac{4}{3}+\tfrac{1}{2})+\cdots+(\tfrac{2}{n-1}-\tfrac{4}{n}+\tfrac{2}{n+1})+(\tfrac{2}{n}-\tfrac{4}{n+1}+\tfrac{2}{n+2})\).

All the interior terms cancel, leaving

\(\sum_{r=1}^{n} \frac{4}{r(r+1)(r+2)}=1-\frac{2}{n+1}+\frac{2}{n+2}.\)

Finally, as \(n\to\infty\), both fractions tend to \(0\), so

\(\sum_{r=1}^{\infty} \frac{4}{r(r+1)(r+2)}=1.\)