Find a formula for the nth partial sum of the series and use it to find the series' sum if the series converges.
-$$\frac { 3 } { 3 \cdot 1 } + \frac { 3 } { 5 \cdot 3 } + \frac { 3 } { 7 \cdot 5 } + \ldots + \frac { 3 } { ( 2 n + 1 ) ( 2 n - 1 ) } + \ldots$$
A) $\frac { 3 n } { 2 ( n + 1 ) } ; \frac { 3 } { 2 }$
B) $\frac { 6 n } { 2 n + 1 } ; 3$
C) $\frac { 3 n } { n + 2 } ; 3$
D) $\frac { 3 n } { 2 n + 1 } ; \frac { 3 } { 2 }$

Question

Quizplus · Accepted Answer

To find the nth partial sum, we need to add up to the nth term of the series. Therefore, we have:

$S_n = \frac{3}{3\cdot1} + \frac{3}{5\cdot3} + \frac{3}{7\cdot 5} + \ldots + \frac{3}{(2n+1)(2n-1)}$

$S_n = 3\left(\frac{1}{3\cdot1} - \frac{1}{5\cdot3} + \frac{1}{5\cdot3} - \frac{1}{7\cdot5} + \ldots + \frac{1}{(2n+1)(2n-1)} - \frac{1}{(2n+3)(2n+1)}ight)$

Observe that the terms in parentheses cancel each other out, and we are left with:

$S_n = \frac{3}{3\cdot1} - \frac{3}{(2n+3)(2n+1)}$

$S_n = 1 - \frac{3}{(2n+3)(2n+1)}$

To find the sum of the series, we need to take the limit as $n$ approaches infinity:

$\lim_{n	o\infty} S_n = \lim_{n	o\infty} \left(1 - \frac{3}{(2n+3)(2n+1)}ight)$

$\lim_{n	o\infty} S_n = \boxed{\frac{3}{2}}$

Find a Formula for the Nth Partial Sum of the Series