Find a formula for the nth partial sum of the series and use it to find the series' sum if the series converges.
-$$\frac { 7 } { 2 \cdot 3 } + \frac { 7 } { 3 \cdot 4 } + \frac { 7 } { 4 \cdot 5 } + \ldots + \frac { 7 } { ( n + 1 ) ( n + 2 ) } + \ldots$$
A) $\frac { 7 n } { n + 2 } ; 7$

B) $\frac { 7 n } { 2 ( n + 1 ) } ; \frac { 7 } { 2 }$

C) $\frac { 7 n } { 2 ( n + 2 ) } ; \frac { 7 } { 2 }$

D) $\frac { 7 n } { n + 1 } ; 7$

Question

Find a formula for the nth partial sum of the series and use it to find the series' sum if the series converges.
-$$\frac { 7 } { 2 \cdot 3 } + \frac { 7 } { 3 \cdot 4 } + \frac { 7 } { 4 \cdot 5 } + \ldots + \frac { 7 } { ( n + 1 ) ( n + 2 ) } + \ldots$$
A) $\frac { 7 n } { n + 2 } ; 7$

B) $\frac { 7 n } { 2 ( n + 1 ) } ; \frac { 7 } { 2 }$

C) $\frac { 7 n } { 2 ( n + 2 ) } ; \frac { 7 } { 2 }$

D) $\frac { 7 n } { n + 1 } ; 7$

Quizplus · Accepted Answer

The given series can be written as: 
$$\frac { 7 } { 2 \cdot 3 } + \frac { 7 } { 3 \cdot 4 } + \frac { 7 } { 4 \cdot 5 } + \ldots + \frac { 7 } { ( n + 1 ) ( n + 2 ) } + \ldots$$
The nth partial sum can be found by summing up to the nth term: 
$$S_n = \frac{7}{2\cdot3} + \frac{7}{3\cdot4} + \frac{7}{4\cdot5} + \ldots + \frac{7}{(n+1)(n+2)}$$
$$= 7\left(\frac{1}{2}\cdot\frac12 + \frac{1}{3}\cdot\frac23 + \frac{1}{4}\cdot\frac34 + \ldots + \frac{1}{n+1}\cdot\frac{n+1}{n+2}\right)$$
$$= \frac{7}{2}\left(1-\frac12 + \frac23 - \frac34 + \ldots + \frac{n}{n+1} - \frac{n+1}{n+2}\right)$$
$$= \frac{7}{2}\left(1 - \frac{1}{n+2}\right)$$
$$= \frac{7n}{2(n+2)}$$

Taking the limit as n approaches infinity, we can see that the sum of the series converges to $\dfrac72$. Hence, the correct choice is (C).

Find a Formula for the Nth Partial Sum of the Series