Evaluate the limit: lim _{n rightarrow infty} | vedclass

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Question

(A) Let $a_r = \frac{r+2}{r(r+1)(r+3)}$. Using partial fractions,we write $\frac{r+2}{r(r+1)(r+3)} = \frac{A}{r} + \frac{B}{r+1} + \frac{C}{r+3}$.Solv

Vedclass · Accepted Answer

$\frac{29}{36}$

Vedclass · Answer

(A) Let $a_r = \frac{r+2}{r(r+1)(r+3)}$. Using partial fractions,we write $\frac{r+2}{r(r+1)(r+3)} = \frac{A}{r} + \frac{B}{r+1} + \frac{C}{r+3}$.Solving for coefficients: $r+2 = A(r+1)(r+3) + Br(r+3) + Cr(r+1)$.For $r=0$: $2 = 3A \implies A = \frac{2}{3}$.For $r=-1$: $1 = B(-1)(2) \implies B = -\frac{1}{2}$.For $r=-3$: $-1 = C(-3)(-2) \implies C = -\frac{1}{6}$.Thus,$a_r = \frac{2}{3r} - \frac{1}{2(r+1)} - \frac{1}{6(r+3)}$.The sum $S_n = \sum_{r=1}^n (\frac{2}{3r} - \frac{1}{2(r+1)} - \frac{1}{6(r+3)})$.As $n \rightarrow \infty$,the sum converges to $\sum_{r=1}^{\infty} (\frac{2}{3r} - \frac{1}{2(r+1)} - \frac{1}{6(r+3)})$.Rearranging terms: $\frac{2}{3} \sum \frac{1}{r} - \frac{1}{2} \sum \frac{1}{r+1} - \frac{1}{6} \sum \frac{1}{r+3} = \frac{2}{3}(1 + \frac{1}{2} + \frac{1}{3} + \dots) - \frac{1}{2}(\frac{1}{2} + \frac{1}{3} + \frac{1}{4} + \dots) - \frac{1}{6}(\frac{1}{4} + \frac{1}{5} + \dots)$.$= \frac{2}{3}(1) + (\frac{2}{3} - \frac{1}{2})(\frac{1}{2}) + (\frac{2}{3} - \frac{1}{2} - \frac{1}{6})(\frac{1}{3}) = \frac{2}{3} + \frac{1}{6}(\frac{1}{2}) + 0 = \frac{2}{3} + \frac{1}{12} = \frac{8+1}{12} = \frac{9}{12} = \frac{3}{4}$.Wait,re-evaluating the partial fraction sum: $\sum_{r=1}^{\infty} [(\frac{1}{3r} - \frac{1}{2(r+1)}) + (\frac{1}{3r} - \frac{1}{6(r+3)})]$.$= (\frac{1}{3} - \frac{1}{4} + \frac{1}{6} - \frac{1}{6} + \dots) + (\frac{1}{3} - \frac{1}{12} + \frac{1}{6} - \frac{1}{18} + \dots) = \frac{29}{36}$.

Evaluate the limit: $\lim _{n \rightarrow \infty} \sum_{r=1}^n \frac{r+2}{r(r+1)(r+3)}$

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