Determine if the series converges or diverges. If the series converges, find its sum.
-$$\sum _ { n = 1 } ^ { \infty } \frac { 3 } { n ( n + 3 ) }$$
A) converges; $\frac { 7 } { 6 }$
B) diverges
C) converges; $\frac { 11 } { 6 }$
D) converges; $\frac { 5 } { 2 }$

Question

Quizplus · Accepted Answer

To determine whether the series converges or diverges, we can use the partial fraction decomposition:

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

Therefore, we have

$$\sum_{n=1}^\infty \frac{3}{n(n+3)}=\sum_{n=1}^\infty \left(\frac{1}{n}-\frac{1}{n+3}ight)$$

Now we can simplify the series:

\begin{align*}
\sum_{n=1}^\infty \left(\frac{1}{n}-\frac{1}{n+3}ight) &= \left(\frac{1}{1}-\frac{1}{4}ight) + \left(\frac{1}{2}-\frac{1}{5}ight) + \left(\frac{1}{3}-\frac{1}{6}ight) + \ldots \
&= 1+\frac{1}{2}+\frac{1}{3}-\left(\frac{1}{4}+\frac{1}{5}+\frac{1}{6}+\ldotsight)
\end{align*}

The alternating harmonic series $\sum_{n=1}^\infty \frac{(-1)^{n-1}}{n}$ converges, so its terms approach 0, meaning that its tail $\frac{1}{4}+\frac{1}{5}+\frac{1}{6}+\ldots$ also approaches 0. Therefore, we have:

$$\sum_{n=1}^\infty \frac{3}{n(n+3)} = 1+\frac{1}{2}+\frac{1}{3}-\left(\frac{1}{4}+\frac{1}{5}+\frac{1}{6}+\ldotsight) = \frac{11}{6}$$

So the series converges and the sum is $\frac{11}{6}$. The answer is C.

Determine If the Series Converges or Diverges ∑n=1∞3n(n+3)\sum _ { n = 1 } ^ { \infty } \frac { 3 } { n ( n + 3 ) }n=1∑∞​n(n+3)3​

Determine If the Series Converges or Diverges $\sum _ { n = 1 } ^ { \infty } \frac { 3 } { n ( n + 3 ) }$