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 term into partial fractions.
\[ \frac{r+2}{r(r+1)(r+3)} = \frac{A}{r}+\frac{B}{r+1}+\frac{C}{r+3} \] Multiply through: $ r+2 = A(r+1)(r+3)+Br(r+3)+Cr(r+1) $.
Step 2: Find A, B, C by substituting roots.
$ r=0 $: $ 2=3A \implies A=\frac{2}{3} $. $ r=-1 $: $ 1=-2B \implies B=-\frac{1}{2} $. $ r=-3 $: $ -1=6C \implies C=-\frac{1}{6} $. Note: $ A+B+C=0 $ (which is needed for convergence).
Step 3: Write $ S_n $ in terms of harmonic numbers.
\[ S_n = \frac{2}{3}H_n - \frac{1}{2}(H_{n+1}-1) - \frac{1}{6}(H_{n+3}-\tfrac{11}{6}) \] where $ H_n = \sum_{r=1}^n \frac{1}{r} $.
Step 4: Observe that divergent harmonic terms cancel.
The coefficients $ \frac{2}{3}-\frac{1}{2}-\frac{1}{6} = 0 $, so all $ H_n $ terms cancel as $ n\to\infty $. Only finite correction terms remain.
Step 5: Collect the finite constant terms from re-indexing.
\[ \lim_{n\to\infty}S_n = \frac{1}{2}\cdot 1 + \frac{1}{6}\cdot\frac{11}{6} = \frac{1}{2}+\frac{11}{36} = \frac{18+11}{36} = \frac{29}{36} \]
Step 6: State the limit.
\[ \boxed{\dfrac{29}{36}} \]