for the following telescoping series, find a formula for the nth term of the sequence of partial sums {s_n}…

Explanation:

Step1: Decompose the general term

Use partial - fraction decomposition. Let $\frac{1}{(4k + 8)(4k+12)}=\frac{A}{4k + 8}+\frac{B}{4k+12}$. Then $1=A(4k + 12)+B(4k + 8)$. Let $4k+8 = 0$ (i.e., $k=-2$), we get $1 = 4A$, so $A=\frac{1}{4}$. Let $4k + 12=0$ (i.e., $k=-3$), we get $1=-4B$, so $B =-\frac{1}{4}$. So $\frac{1}{(4k + 8)(4k+12)}=\frac{1}{4}\left(\frac{1}{4k + 8}-\frac{1}{4k+12}\right)$.

Step2: Find the $n$th partial sum $S_n$

$S_n=\sum_{k = 0}^{n}\frac{1}{(4k + 8)(4k+12)}=\frac{1}{4}\sum_{k = 0}^{n}\left(\frac{1}{4k + 8}-\frac{1}{4k+12}\right)$. Expanding the sum: When $k = 0$, the term is $\frac{1}{4}\left(\frac{1}{8}-\frac{1}{12}\right)$; when $k = 1$, the term is $\frac{1}{4}\left(\frac{1}{12}-\frac{1}{16}\right)$; $\cdots$; when $k=n$, the term is $\frac{1}{4}\left(\frac{1}{4n + 8}-\frac{1}{4n+12}\right)$. By telescoping, $S_n=\frac{1}{4}\left(\frac{1}{8}-\frac{1}{4n + 12}\right)$.

Step3: Evaluate the limit

$\lim_{n\rightarrow\infty}S_n=\lim_{n\rightarrow\infty}\frac{1}{4}\left(\frac{1}{8}-\frac{1}{4n + 12}\right)$. As $n\rightarrow\infty$, $\frac{1}{4n + 12}\rightarrow0$. So $\lim_{n\rightarrow\infty}S_n=\frac{1}{32}$.

Answer:

$S_n=\frac{1}{4}\left(\frac{1}{8}-\frac{1}{4n + 12}\right)$; the value of the series is $\frac{1}{32}$