Sum $\sum_{n=1}^{\infty}\frac{3n+7}{n(n+1)(n+2)}.$

Question

How can we find the sum

$$\sum_{n=1}^{\infty}\frac{3n+7}{n(n+1)(n+2)}.$$

WolframAlpha says it's $13/4$ but how to compute it? In general how to approach this kind of sums of rational functions where denominator has distinct roots?

Answer 1

$$\frac{3n+7}{n(n+1)(n+2)}=\frac{3n+6+1}{n(n+1)(n+2)}=$$ $$=\frac{3}{n(n+1)}+\frac{1}{n(n+1)(n+2)}=\frac{3}{n}-\frac{3}{n+1}+\frac{1}{n}\left(\frac{1}{n+1}-\frac{1}{n+2}\right)=$$ $$=\frac{3}{n}-\frac{3}{n+1}+\frac{1}{n}-\frac{1}{n+1}-\frac{1}{2}\left(\frac{1}{n}-\frac{1}{n+2}\right)=$$ $$=\frac{7}{2n}-\frac{4}{n+1}+\frac{1}{2(n+2)}$$ and use the telescopic sum

Answer 2

Using partial fractions we have \begin{eqnarray*} \frac{3n+7}{n(n+1)(n+2)}= \frac{\frac{7}{2}}{n}+\frac{-4}{n+1}+\frac{\frac{1}{2}}{n+2} \end{eqnarray*} It is a telescoping sum ... \begin{eqnarray*} \sum_{n=1}^{\infty} \frac{3n+7}{n(n+1)(n+2)} = \frac{\frac{7}{2}}{1}&+&\color{green}{\frac{-4}{2}} &+&\color{red}{\frac{\frac{1}{2}}{3}} \\&+&\color{green}{\frac{\frac{7}{2}}{2}}&+&\color{red}{\frac{-4}{3}+\frac{\frac{1}{2}}{4}} \\ && &+&\color{red}{\frac{\frac{7}{2}}{3}}+\color{red}{\frac{-4}{4}+\frac{\frac{1}{2}}{5}} \\ &&&& \ddots \end{eqnarray*} After the dust settles ... we have $\color{red}{\frac{13}{4}}$.

Answer 3

$S=\sum_{n=1}^{+\infty}(\frac {3}{(n+1)(n+2)}+\frac {7}{n (n+1)(n+2)}) $

$$=3\sum_{n=1}^{+\infty}(\frac {1}{n+1}-\frac {1}{n+2}) $$

$$+7\sum_{n=1}^{+\infty}\frac{1}{n (n+1)(n+2)} $$

$$\implies S=\frac {3}{2}+7A$$

and

$$S=\sum_{n=1}^{+\infty}\frac {3 (n+2)+1}{n (n+1)(n+2)} $$ $$=3\sum_{n=1}^{+\infty}(\frac {1}{n}-\frac {1}{n+1})+A$$

$$\implies S=3+A $$

thus after eliminating $A $, we get $$6S=21-\frac {3}{2}\implies S=\frac {13}{4} .$$

Answer 4

Alternatively:

$$\sum_{n=1}^{\infty}\frac{3n+7}{n(n+1)(n+2)}=\sum_{n=1}^{\infty}\left(\frac{4}{n(n+1)}-\frac{1}{n(n+2)}\right)=$$

$$4\sum_{n=1}^{\infty}\left(\frac{1}{n}-\frac{1}{n+1}\right)-\frac 12\sum_{n=1}^{\infty}\left(\frac{1}{n}-\frac{1}{n+2}\right)=$$

$$4\left(1-\require{cancel}\cancel{\frac 12}+\cancel{\frac12}-\cancel{\frac13}+\cancel{\frac13}-\cancel{\frac14}+...\right)-\frac12\left(1-\cancel{\frac13}+\frac12-\cancel{\frac14}+\cancel{\frac13}-\cancel{\frac15}+...\right)=$$ $$4-\frac{1}{2}\cdot\frac32=\frac{13}{4}.$$

Answer 5

As an alternative approach, since by computing residues we have $$ \frac{3n+7}{n(n+1)(n+2)} = \frac{7}{2}\cdot\frac{1}{n}-4\cdot\frac{1}{n+1}+\frac{1}{2}\cdot\frac{1}{n+2} \tag{1}$$ it follows that $$ S=\sum_{n\geq 1}\frac{3n+7}{n(n+1)(n+2)} = \int_{0}^{1}\left(\frac{7}{2}-4x+\frac{x^2}{2}\right)\sum_{n\geq 1}x^{n-1}\,dx \tag{2}$$ hence: $$ S = \frac{1}{2}\int_{0}^{1}\frac{7-8x+x^2}{1-x}\,dx = \frac{1}{2}\int_{0}^{1}(7-x)\,dx = \frac{1}{2}\left(7-\frac{1}{2}\right)=\color{red}{\frac{13}{4}}.\tag{3} $$