This is a telescoping series whose sum is
$$\sum_{i=1}^{\infty} \frac{6}{n(n+3)} = \frac{11}{3}$$
I calculated it as $$\left(\frac{2}{1} - \frac{2}{1+3}\right) - \left(\frac{2}{\infty} - \frac{2}{\infty+3}\right) = 2 - \frac{2}{4} = \frac{3}{2}$$
However, the answer was $$2\left(1+\frac{1}{2}+ \frac{1}{3}\right)= \frac{11}{3}$$ How was this calculated? It seems like they just added a few numbers out of the whole series.
On
Whenever we are dealing with infinite series, we should always look at it's partial sums. Consider $N \in \mathbb{N}$, and notice that $$S_{N} = \sum_{n=1}^{N} \frac{6}{n(n+3)} = 2\sum_{n=1}^{N} \frac{(n+3)-n}{n(n+3)} = 2\left(1+\frac{1}{2}+\frac{1}{3}-\frac{1}{N+1}-\frac{1}{N+2}-\frac{1}{N+3}\right).$$ I think you have just missed a few terms while computing the partial sums.
On
In the language of OP, the correct answer is $$ 2+\frac{2}{1+1}+\frac{2}{1+2}-\left(\frac{2}{\infty}+\frac{2}{\infty+1}+ +\frac{2}{\infty+1}\right)=\frac{11}{3} $$
$$ \begin{aligned} \sum_{n=1}^{\infty} \frac{3}{n(n+1)} &=\sum_{n=1}^{\infty}\left(\frac{1}{n}-\frac{1}{n+3}\right) \\ &=\lim _{N \rightarrow \infty}\left(\sum_{n=1}^{N} \frac{1}{n}-\sum_{n=1}^{N} \frac{1}{n+3}\right) \\ &=\lim _{N \rightarrow \infty}\left[ \sum_{n=1}^{N} \frac{1}{n}-\left(\sum_{n=1}^{N} \frac{1}{n}+\frac{1}{N+3}+\frac{1}{N+2}+\frac{1}{N+1}-1-\frac{1}{2}-\frac{1}{3}\right)\right]\\ &=\lim _{N \rightarrow \infty}\left[ -\frac{1}{N+3}-\frac{1}{N+2}-\frac{1}{N+1}+1+\frac{1}{2}+\frac{1}{3}\right]\\&=\frac{11}{6} \end{aligned} $$ Hence $$ \sum_{n=1}^{\infty} \frac{6}{n(n+1)}=\frac{11}{3} $$
On
More generally, if
$s(k) =\sum_{n=1}^{\infty} \dfrac1{n(n+k)} $, consider the partial sum $s(k, m) =\sum_{n=1}^{m} \dfrac1{n(n+k)} $ where $m > k$.
Then $s(k) =\lim_{m \to \infty} s(k, m) =\dfrac1{k} \sum_{n=1}^{k} \dfrac1{n} $.
Note that $s(k) =\dfrac1{k}H_k \approx \dfrac{\ln(k)}{k} $.
Proof.
If $m > k$ then
$\begin{array}\\ s(k, m) &=\sum_{n=1}^{m} \dfrac1{n(n+k)}\\ &=\sum_{n=1}^{m} \dfrac1{k}\left(\dfrac1{n}-\dfrac1{n+k}\right)\\ &=\dfrac1{k}\sum_{n=1}^{m} \left(\dfrac1{n}-\dfrac1{n+k}\right)\\ &=\dfrac1{k}\left(\sum_{n=1}^{m} \dfrac1{n}-\sum_{n=1}^{m} \dfrac1{n+k}\right)\\ &=\dfrac1{k}\left(\sum_{n=1}^{m} \dfrac1{n}-\sum_{n=k+1}^{k+m} \dfrac1{n}\right)\\ &=\dfrac1{k}\left(\sum_{n=1}^{k} \dfrac1{n}+\sum_{n=k+1}^{m} \dfrac1{n}-\sum_{n=k+1}^{m} \dfrac1{n}-\sum_{n=m+1}^{k+m} \dfrac1{n}\right)\\ &=\dfrac1{k}\left(\sum_{n=1}^{k} \dfrac1{n}-\sum_{n=m+1}^{k+m} \dfrac1{n}\right)\\ d(k, m) &=s(k,m)-\dfrac1{k} \sum_{n=1}^{k} \dfrac1{n}\\ &=-\dfrac1{k}\sum_{n=m+1}^{k+m} \dfrac1{n}\\ |d(k, m)| &=\dfrac1{k}\sum_{n=m+1}^{k+m} \dfrac1{n}\\ &\le\dfrac1{k}\sum_{n=m+1}^{k+m} \dfrac1{m+1}\\ &=\dfrac1{k}\sum_{n=m+1}^{k+m} \dfrac1{m+1}\\ &=\dfrac1{k} \dfrac{k}{m+1}\\ &=\dfrac1{m+1}\\ &\to 0\\ \end{array} $
On
Generalizing Marty Cohen's answer, and assuming $0\le a<b$, if $$s(a,b)=\sum_{n=1}^{\infty} \dfrac1{(n+a)(n+b)}, $$ consider the partial sum $$s(a, b, m) =\sum_{n=1}^{m} \dfrac1{(n+a)(n+b)}, \text{where $m>b$} $$
Then $$s(a,b) =\lim_{m \to \infty} s(a, b, m) =\dfrac1{b-a} \sum_{n=a+1}^{b} \dfrac1{n} $$
Proof.
If $m > b$ then
$$\begin{array}\\ s(a, b, m) &=\sum_{n=1}^{m} \dfrac1{(n+a)(n+b)}\\ &=\sum_{n=1}^{m} \dfrac1{b-a}\left(\dfrac1{n+a}-\dfrac1{n+b}\right)\\ &=\dfrac1{b-a}\left(\sum_{n=1}^{m} \dfrac1{n+a}-\sum_{n=1}^{m} \dfrac1{n+b}\right)\\ &=\dfrac1{b-a}\left(\sum_{n=a+1}^{a+m} \dfrac1{n}-\sum_{n=b+1}^{b+m} \dfrac1{n}\right)\\ &=\dfrac1{b-a}\left(\sum_{n=a+1}^{b} \dfrac1{n}+\sum_{n=b+1}^{a+m} \dfrac1{n}-\sum_{n=b+1}^{a+m} \dfrac1{n}-\sum_{n=a+m+1}^{b+m} \dfrac1{n}\right)\\ &=\dfrac1{b-a}\left(\sum_{n=a+1}^{b} \dfrac1{n}-\sum_{n=a+m+1}^{b+m} \dfrac1{n}\right)\\ d(a, b, m) &=s(a,b,m)-\dfrac1{b-a}\sum_{n=a+1}^{b} \dfrac1{n}\\ &=-\dfrac1{b-a}\sum_{n=a+m+1}^{b+m} \dfrac1{n}\\ |d(a,b, m)| &=\dfrac1{b-a}\sum_{n=a+m+1}^{b+m} \dfrac1{n}\\ &\le\dfrac1{b-a}\sum_{n=a+m+1}^{b+m} \dfrac1{a+m+1}\\ &=\dfrac1{b-a} \cdot\dfrac{b-a}{a+m+1}\\ &=\dfrac1{a+m+1}\\ &{\to 0} \ \ \text{for $m\to +\infty$}\\ \end{array} $$ Examples: $$\color{red}{s(a,b) =\sum_{n=1}^{\infty} \dfrac1{(n+a)(n+b)} =\dfrac1{b-a} \sum_{n=a+1}^{b} \dfrac1{n} }\\ s(0,3)=\sum_{n=1}^{\infty} \dfrac1{n(n+3)}=\dfrac13 \sum_{n=1}^{3} \dfrac1{n}=\frac{11}{18}\\ s(1,3)=\sum_{n=1}^{\infty} \dfrac1{(n+1)(n+3)}=\dfrac12 \sum_{n=2}^{3} \dfrac1{n}=\frac{5}{12}\\ $$
Solution:
Since $\forall k\in \mathbb{Z}^+$, there holds that
$$ \frac{6}{k(k+3)}=2\left(\frac{1}{k}-\frac{1}{k+3}\right), $$
we can calculate the partial sum
\begin{align} &\sum_{k=1}^n\frac{6}{k(k+3)}\\ =~&2\sum_{k=1}^n\left(\frac{1}{k}-\frac{1}{k+3}\right)\\ =~&2\left[\left(1-\frac{1}{4}\right)+\left(\frac{1}{2}-\frac{1}{5}\right)+\left(\frac{1}{3}-\frac{1}{6}\right)+\cdots+\left(\frac{1}{n-2}-\frac{1}{n+1}\right)\\ +\left(\frac{1}{n-1}-\frac{1}{n+2}\right)+\left(\frac{1}{n}-\frac{1}{n+3}\right)\right]\\ =~&2\left(1+\frac{1}{2}+\frac{1}{3}-\frac{1}{n+1}-\frac{1}{n+2}-\frac{1}{n+3}\right). \end{align}
Hence, the original infinite sum is
\begin{align} &\sum_{k=1}^\infty\frac{6}{k(k+3)}\\ =~&\underset{n\rightarrow \infty}{\lim} 2\left(1+\frac{1}{2}+\frac{1}{3}-\frac{1}{n+1}-\frac{1}{n+2}-\frac{1}{n+3}\right)\\ =~&2\left(1+\frac{1}{2}+\frac{1}{3}\right)\\ =~&\frac{11}{3}. \end{align}