Find a formula for the nth partial sum of the series and use it to find the series' sum if the series converges.
-$$\frac { 8 } { 1 \cdot 3 } + \frac { 8 } { 2 \cdot 4 } + \frac { 8 } { 3 \cdot 5 } + \ldots + \frac { 8 } { n ( n + 2 ) } + \ldots$$
A) $\frac { 8 n ( 3 n + 4 ) } { 4 n ( n + 1 ) } ; 6$

B) $\frac { 8 n ( 3 n + 5 ) } { 4 ( n + 1 ) ( n + 2 ) } ; 6$

C) $\frac { 8 n ( 3 n + 5 ) } { 4 n ( n + 2 ) } ; 6$

D) $\frac { 8 n ( 3 n + 4 ) } { 4 n ( n + 2 ) } ; 6$

Question

Find a formula for the nth partial sum of the series and use it to find the series' sum if the series converges.
-$$\frac { 8 } { 1 \cdot 3 } + \frac { 8 } { 2 \cdot 4 } + \frac { 8 } { 3 \cdot 5 } + \ldots + \frac { 8 } { n ( n + 2 ) } + \ldots$$
A) $\frac { 8 n ( 3 n + 4 ) } { 4 n ( n + 1 ) } ; 6$

B) $\frac { 8 n ( 3 n + 5 ) } { 4 ( n + 1 ) ( n + 2 ) } ; 6$

C) $\frac { 8 n ( 3 n + 5 ) } { 4 n ( n + 2 ) } ; 6$

D) $\frac { 8 n ( 3 n + 4 ) } { 4 n ( n + 2 ) } ; 6$

Quizplus · Accepted Answer

The given series can be rewritten using partial fraction decomposition. The general term $\frac{8}{n(n+2)}$ can be decomposed into $\frac{A}{n} + \frac{B}{n+2}$, where $A$ and $B$ are constants to be determined. Solving for $A$ and $B$, we find $A = 4$ and $B = -4$, making the term equivalent to $\frac{4}{n} - \frac{4}{n+2}$. This structure suggests a telescoping series, where most terms cancel out, leaving only the first few and the last few terms. Summing up to the nth term and taking the limit as $n$ approaches infinity (for the series sum) reveals that the series converges to 6. The correct formula for the nth partial sum that matches this process is $\frac { 8 n ( 3 n + 5 ) } { 4 ( n + 1 ) ( n + 2 ) }$, and the series converges to 6, which corresponds to option B.

Find a Formula for the Nth Partial Sum of the Series