4. $sum_{n = 1}^{infty}(\frac{9}{(n + 9)(n + 11)})=$\na. $\frac{189}{220}$\nb. $\frac{135}{143}$\nc…

Explanation:

Step1: Decompose the fraction

Use partial - fraction decomposition. Let $\frac{9}{(n + 9)(n+11)}=\frac{A}{n + 9}+\frac{B}{n + 11}$. Then $9=A(n + 11)+B(n + 9)$. Let $n=-9$, we get $9 = 2A$, so $A=\frac{9}{2}$. Let $n=-11$, we get $9=-2B$, so $B =-\frac{9}{2}$. So $\frac{9}{(n + 9)(n+11)}=\frac{9}{2}\left(\frac{1}{n + 9}-\frac{1}{n + 11}\right)$.

Step2: Write out the series

The series $\sum_{n = 1}^{\infty}\frac{9}{(n + 9)(n+11)}=\frac{9}{2}\sum_{n = 1}^{\infty}\left(\frac{1}{n + 9}-\frac{1}{n + 11}\right)$.

Step3: Expand the series

When $n = 1$, the term is $\frac{1}{10}-\frac{1}{12}$; when $n = 2$, the term is $\frac{1}{11}-\frac{1}{13}$; when $n = 3$, the term is $\frac{1}{12}-\frac{1}{14}$; $\cdots$; when $n=k$, the term is $\frac{1}{k + 9}-\frac{1}{k + 11}$.

Step4: Find the sum of the series

This is a telescoping series. After cancellation, the sum $S=\frac{9}{2}\left(\frac{1}{10}+\frac{1}{11}\right)$.

Step5: Calculate the result

$S=\frac{9}{2}\times\frac{11 + 10}{110}=\frac{9}{2}\times\frac{21}{110}=\frac{189}{220}$.

Answer:

A. $\frac{189}{220}$