determine whether the series is convergent or divergent by expressing ( s_{n} ) as a telescoping sum. if it…

Explanation:

Step1: Decompose the general term

Use partial - fraction decomposition. Let (\frac{27}{n(n + 3)}=\frac{A}{n}+\frac{B}{n + 3}). Then (27=A(n + 3)+Bn=(A + B)n+3A). Solving the system (\begin{cases}A + B = 0\3A=27\end{cases}), we get (A = 9) and (B=-9). So (\frac{27}{n(n + 3)}=\frac{9}{n}-\frac{9}{n + 3}).

Step2: Find the (n) - th partial sum (s_n)

(s_n=\sum_{k = 1}^{n}\frac{27}{k(k + 3)}=\sum_{k = 1}^{n}\left(\frac{9}{k}-\frac{9}{k + 3}\right)) (s_n=9\left[\left(1-\frac{1}{4}\right)+\left(\frac{1}{2}-\frac{1}{5}\right)+\left(\frac{1}{3}-\frac{1}{6}\right)+\left(\frac{1}{4}-\frac{1}{7}\right)+\cdots+\left(\frac{1}{n-2}-\frac{1}{n + 1}\right)+\left(\frac{1}{n-1}-\frac{1}{n + 2}\right)+\left(\frac{1}{n}-\frac{1}{n + 3}\right)\right]) After cancellation (telescoping), (s_n=9\left(1+\frac{1}{2}+\frac{1}{3}-\frac{1}{n + 1}-\frac{1}{n + 2}-\frac{1}{n + 3}\right))

Step3: Find the limit of (s_n) as (n\to\infty)

(\lim_{n\to\infty}s_n=\lim_{n\to\infty}9\left(1+\frac{1}{2}+\frac{1}{3}-\frac{1}{n + 1}-\frac{1}{n + 2}-\frac{1}{n + 3}\right)) Since (\lim_{n\to\infty}\frac{1}{n + 1}=\lim_{n\to\infty}\frac{1}{n + 2}=\lim_{n\to\infty}\frac{1}{n + 3}=0) (\lim_{n\to\infty}s_n=9\left(1+\frac{1}{2}+\frac{1}{3}\right)=9\times\frac{6 + 3+2}{6}=9\times\frac{11}{6}=\frac{33}{2})

Answer:

(\frac{33}{2})