Question

Evaluate \[ \lim_{n\to \infty}\sum_{r=1}^{n}\frac{r+2}{r(r+1)(r+3)} \] is equal to:

• A. \(\dfrac{29}{36}\)
• B. \(\dfrac{1}{36}\)
• C. \(\dfrac{5}{36}\)
• D. \(\dfrac{23}{36}\)

Hint:

For infinite sums involving rational functions of \(r\), use partial fractions first and then check cancellation of harmonic terms.

Answer 1

The Correct Option is A

Step 1: Decompose the given fraction.
We have \[ \frac{r+2}{r(r+1)(r+3)} \] Let \[ \frac{r+2}{r(r+1)(r+3)} = \frac{A}{r}+\frac{B}{r+1}+\frac{C}{r+3} \] Multiplying by \(r(r+1)(r+3)\), \[ r+2=A(r+1)(r+3)+Br(r+3)+Cr(r+1) \] Putting \(r=0\): \[ 2=3A \] \[ A=\frac{2}{3} \] Putting \(r=-1\): \[ 1=-2B \] \[ B=-\frac{1}{2} \] Putting \(r=-3\): \[ -1=6C \] \[ C=-\frac{1}{6} \] Therefore, \[ \frac{r+2}{r(r+1)(r+3)} = \frac{2}{3r}-\frac{1}{2(r+1)}-\frac{1}{6(r+3)} \]

Step 2: Write the infinite sum.
\[ \sum_{r=1}^{\infty}\frac{r+2}{r(r+1)(r+3)} = \sum_{r=1}^{\infty} \left[ \frac{2}{3r}-\frac{1}{2(r+1)}-\frac{1}{6(r+3)} \right] \]

Step 3: Combine harmonic terms carefully.
The \(n\)-th partial sum is \[ S_n= \frac{2}{3}\sum_{r=1}^{n}\frac{1}{r} -\frac{1}{2}\sum_{r=1}^{n}\frac{1}{r+1} -\frac{1}{6}\sum_{r=1}^{n}\frac{1}{r+3} \] Using harmonic numbers, \[ S_n= \frac{2}{3}H_n-\frac{1}{2}(H_{n+1}-1)-\frac{1}{6}(H_{n+3}-H_3) \] Since \[ H_3=1+\frac12+\frac13=\frac{11}{6}, \] we get \[ S_n= \frac{2}{3}H_n-\frac{1}{2}H_{n+1}+\frac12-\frac{1}{6}H_{n+3}+\frac{11}{36} \]

Step 4: Take the limit.
As \(n\to\infty\), the divergent harmonic terms cancel because \[ \frac{2}{3}-\frac{1}{2}-\frac{1}{6}=0 \] So, \[ \lim_{n\to\infty}S_n = \frac12+\frac{11}{36} \] \[ = \frac{18}{36}+\frac{11}{36} \] \[ = \frac{29}{36} \]

Step 5: Final conclusion.
Therefore, \[ \boxed{\frac{29}{36}} \]