Calculate the following convergent series: $\sum _{n=1}^{\infty }\:\frac{1}{n\left(n+3\right)}$

Question

I need to tell if a following series convergent and if so, find it's value: $$ \sum _{n=1}^{\infty }\:\frac{1}{n\left(n+3\right)} $$

I've noticed that $$ \sum _{n=1}^{\infty }\:\frac{1}{n\left(n+3\right)} = \frac{1}{3}\sum _{n=1}^{\infty }\:\frac{1}{n}-\frac{1}{n+3} = \frac{1}{3}\left(1-\frac{1}{4}+\frac{1}{2}-\frac{1}{5}+\frac{1}{3}-\frac{1}{6}+\frac{1}{4}-\frac{1}{7}+\cdots\right) $$ Wich means some values are zeroed, does that mean it's a telescoping sum?

I also know that $\sum _{n=1}^{\infty }\:\frac{1}{n\left(n+3\right)} \le \sum _{n=1}^{\infty }\:\frac{1}{n^2}\:$ so the series converges by the harmonic p-series.

Answer 1

Yes, you are correct that the sum telescopes quite nicely. The partial sum formula is$$\sum\limits_{n=1}^m\frac 1{n(n+3)}=\frac 13\left[\left(1+\frac 12+\cdots+\frac 1m\right)-\left(\frac 14+\frac 15+\cdots+\frac 1{m+3}\right)\right]$$Notice how anything past $\frac 14$ in the first sum is automatically canceled from the second sum. Hence, the partial sum formula is given as$$\sum\limits_{n=1}^m\frac 1{n(n+3)}=\frac 13\left(1+\frac 12+\frac 13-\frac 1{m+1}-\frac 1{m+2}-\frac 1{m-3}\right)$$As $m\to\infty$, the fractions containing $m$ vanish, leaving$$\sum\limits_{n\geq1}\frac 1{n(n+3)}\color{blue}{=\frac {11}{18}}$$

Answer 2

Your inequality is correct, it does converge to a value and yes, it is telescoping.

Begin by separating the fraction in the sum into two terms with partial fractions

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

Answer 3

Yes, it is telescoping,

\begin{align} \sum_{n=1}^N \frac{1}{n(n+3)}&= \frac13 \sum_{n=1}^N \left[ \frac1{n}-\frac{1}{n+3}\right]\\ &= \frac13\left[1-\frac{1}{4} +\frac12-\frac1{5}+\frac13-\frac16+\ldots +\frac1{N-2}-\frac1{N+1}+\frac1{N-1}-\frac1{N+2}+\frac1N-\frac1{N+3}\right]\\ &=\frac13\left[1+\frac12+\frac13-\frac1{N+1}-\frac{1}{N+2}-\frac1{N+3} \right] \end{align}

Now take limit $N \to \infty$, $$\sum_{n=1}^N \frac{1}{n(n+3)}=\frac13\left[1+\frac12+\frac13 \right]$$

Answer 4

For any positive integer $k$, if $N > k$,

$\begin{array}\\ \sum _{n=1}^{N}\:\frac{1}{n\left(n+k\right)} &=\sum _{n=1}^{N}\frac1{k}\left(\frac1{n}-\frac1{n+k}\right)\\ &=\frac1{k}\left(\sum_{n=1}^{N}\frac1{n}-\sum_{n=1}^{N}\frac1{n+k}\right)\\ &=\frac1{k}\left(\left(\sum _{n=1}^{k}\frac1{n}+\sum _{n=k+1}^{N}\frac1{n}\right)-\left(\sum_{n=k+1}^{k+N}\frac1{n}\right)\right)\\ &=\frac1{k}\left(\left(\sum _{n=1}^{k}\frac1{n}+\sum _{n=k+1}^{N}\frac1{n}\right)-\left(\sum_{n=k+1}^{N}\frac1{n}+\sum_{n=N+1}^{N+k}\frac1{n}\right)\right)\\ &=\frac1{k}\left(\left(\sum_{n=1}^{k}\frac1{n}\right)-\left(\sum_{n=N+1}^{N+k}\frac1{n}\right)\right)\\ &\to \frac1{k}\sum_{n=1}^{k}\frac1{n} \qquad\text{as } N \to \infty \text{ since } \frac{k}{N+k}\le \sum_{n=N+1}^{N+k}\frac1{n} \le \frac{k}{N+1}\\ \end{array} $

Your case is $k=3$.

Answer 5

To show that the series is convergent it sufficies to observe that $$ \lim_{n\to\infty}\frac{1/[n(n+3)]}{1/n^2}=1 $$ and since $0\le\sum_n n^{-2}<\infty$, the series in question is convergent by the limit comparison test.