Determine if the series converges or diverges. If the series converges, find its sum.
-$$\sum _ { n = 1 } ^ { \infty } \frac { 7 } { ( 4 n - 1 ) ( 4 n + 3 ) }$$
A) diverges
B) converges; $\frac { 7 } { 12 }$
C) converges; $\frac { 7 } { 18 }$
D) converges; $\frac { 7 } { 6 }$

Question

Quizplus · Accepted Answer

We can use partial fractions to write 
$$\frac{7}{(4n-1)(4n+3)}=\frac{1}{4n-1}-\frac{1}{4n+3}.$$
Thus,
\begin{align*}
\sum_{n=1}^\infty \frac{7}{(4n-1)(4n+3)}&=\sum_{n=1}^\infty \left(\frac{1}{4n-1}-\frac{1}{4n+3}\right) \
&=\left(\frac{1}{3}-\frac{1}{7}\right)+\left(\frac{1}{7}-\frac{1}{11}\right)+\left(\frac{1}{11}-\frac{1}{15}\right)+\cdots
\end{align*}
The telescoping sum collapses and gives $\sum _ { n = 1 } ^ { \infty } \frac { 7 } { ( 4 n - 1 ) ( 4 n + 3 ) } = \boxed{\frac { 7 } { 12 }}$.

Determine If the Series Converges or Diverges ∑n=1∞7(4n−1)(4n+3)\sum _ { n = 1 } ^ { \infty } \frac { 7 } { ( 4 n - 1 ) ( 4 n + 3 ) }n=1∑∞​(4n−1)(4n+3)7​

Determine If the Series Converges or Diverges $\sum _ { n = 1 } ^ { \infty } \frac { 7 } { ( 4 n - 1 ) ( 4 n + 3 ) }$