Find a formula for the nth partial sum of the series and use it to find the series' sum if the series converges.
-$$\frac { 9 } { 1 ^ { 2 } \cdot 2 ^ { 2 } } + \frac { 15 } { 2 ^ { 2 } \cdot 3 ^ { 2 } } + \frac { 21 } { 3 ^ { 2 } \cdot 4 ^ { 2 } } + \ldots + \frac { 3 ( 2 n + 1 ) } { n ^ { 2 } ( n + 1 ) ^ { 2 } } + \ldots$$
A) $\frac { 3 n ( n + 2 ) } { ( n + 1 ) ^ { 2 } } ; 3$

B) $\frac { 3 n ( n + 1 ) } { ( n + 2 ) ^ { 2 } } ; 3$

C) $\frac { 3 n ^ { 2 } } { ( n + 1 ) ( n + 2 ) } ; 3$

D) $\frac { 3 ( n + 1 ) ( n + 2 ) } { n ^ { 2 } } ; 3$

Question

Find a formula for the nth partial sum of the series and use it to find the series' sum if the series converges.
-$$\frac { 9 } { 1 ^ { 2 } \cdot 2 ^ { 2 } } + \frac { 15 } { 2 ^ { 2 } \cdot 3 ^ { 2 } } + \frac { 21 } { 3 ^ { 2 } \cdot 4 ^ { 2 } } + \ldots + \frac { 3 ( 2 n + 1 ) } { n ^ { 2 } ( n + 1 ) ^ { 2 } } + \ldots$$
A) $\frac { 3 n ( n + 2 ) } { ( n + 1 ) ^ { 2 } } ; 3$

B) $\frac { 3 n ( n + 1 ) } { ( n + 2 ) ^ { 2 } } ; 3$

C) $\frac { 3 n ^ { 2 } } { ( n + 1 ) ( n + 2 ) } ; 3$

D) $\frac { 3 ( n + 1 ) ( n + 2 ) } { n ^ { 2 } } ; 3$

Quizplus · Accepted Answer

The given series can be written as a telescoping series by partial fraction decomposition. Each term can be decomposed as $\frac{3(2n+1)}{n^2(n+1)^2} = \frac{A}{n^2} + \frac{B}{(n+1)^2}$, where $A$ and $B$ are constants to be determined. Solving for $A$ and $B$ gives $A = 3$ and $B = -3$, leading to a telescoping series where most terms cancel out, leaving only the first few and last terms. The nth partial sum simplifies to $\frac{3n(n+2)}{(n+1)^2}$, and as $n$ approaches infinity, the series converges to 3.

Find a Formula for the Nth Partial Sum of the Series