Find the sum of the series $\sum _ { n = 1 } ^ { \infty } \frac { 1 } { n ( n + 2 ) }$ .
A) $\frac { 3 } { 4 }$
B) $\frac { 1 } { 2 }$
C) $\frac { 3 } { 5 }$
D) $\frac { 9 } { 10 }$
E) $\frac { 7 } { 10 }$
F) $\frac { 4 } { 5 }$
G) $\frac { 2 } { 3 }$
H)Divergent

Question

Find the sum of the series $\sum _ { n = 1 } ^ { \infty } \frac { 1 } { n ( n + 2 ) }$ .
A) $\frac { 3 } { 4 }$ 
B) $\frac { 1 } { 2 }$ 
C) $\frac { 3 } { 5 }$ 
D) $\frac { 9 } { 10 }$ 
E) $\frac { 7 } { 10 }$ 
F) $\frac { 4 } { 5 }$ 
G) $\frac { 2 } { 3 }$ 
H)Divergent

Quizplus · Accepted Answer

To find the sum of the series $\sum _ { n = 1 } ^ { \infty } \frac { 1 } { n ( n + 2 ) }$, we can use partial fraction decomposition. The given series can be decomposed as follows:$\frac { 1 } { n ( n + 2 ) } = \frac { A } { n } + \frac { B } { n + 2 }$Solving for A and B, we get $A = \frac { 1 } { 2 }$ and $B = -\frac { 1 } { 2 }$. Thus, the series becomes:$\sum _ { n = 1 } ^ { \infty } \left( \frac { 1 } { 2n } - \frac { 1 } { 2(n + 2) } \right)$This series telescopes, meaning most terms cancel out, leaving only the first few and the last few terms. Specifically, the terms for $n = 1$ and $n = 2$ do not get canceled out:$\frac { 1 } { 2 } - \frac { 1 } { 4 } + \frac { 1 } { 4 } - \frac { 1 } { 6 } + \frac { 1 } { 6 } - \frac { 1 } { 8 } + \cdots = \frac { 1 } { 2 }$Thus, the sum of the series is $\frac { 1 } { 2 } + \frac { 1 } { 4 } = \frac { 3 } { 4 }$.

Find the Sum of the Series ∑n=1∞1n(n+2)\sum _ { n = 1 } ^ { \infty } \frac { 1 } { n ( n + 2 ) }∑n=1∞​n(n+2)1​

Find the Sum of the Series $\sum _ { n = 1 } ^ { \infty } \frac { 1 } { n ( n + 2 ) }$