Exact value of a convergent series: $\sum_{n=1}^\infty \frac{1}{n^3+6n^2+8n}$

Question

I have a series $$\sum_{n=1}^\infty \frac{1}{n^3+6n^2+8n}.$$ I know the series converges because $$\frac{1}{n^3+6n^2+8n}\le \frac{1}{n^3}, $$ since $p=3>1$, I know that $\sum 1/n^3$ converges. But I am not sure how to figure out what it converges to.

Answer 1

Note that $$n^3+6n^2+8n = n(n+4)(n+2)$$

Thus you need to apply partial fraction on $$ \frac {1}{n(n+4)(n+2)} $$and make it some sort of telescoping series to find the exact value.

Answer 2

Note $$ \frac{1}{n^3+6n^2+8n}=\frac1{n(n+2)(n+4)}=\frac14\bigg[\frac{1}{n(n+2)}-\frac{1}{(n+2)(n+4)}\bigg]. $$ Let $$ f(x)=\sum_{n=1}^\infty\frac{1}{n(n+2)}x^n, g(x)=\sum_{n=1}^\infty\frac{1}{(n+2)(n+4)}x^{n+2}. $$ Then \begin{eqnarray} f(1)&=&\int_0^1\frac1{x^3}\int_0^x\frac{t^2}{1-t}dtdx\\ &=&\int_0^1\int_0^x\frac{t^4}{x^2(1-t)}dtdx\\ &=&\int_0^1\int_t^1\frac{t^4}{x^2(1-t)}dxdt\\ &=&\frac12\int_0^1\frac{t^2}{1-t}\bigg(\frac1{t^2}-1\bigg)dt\\ &=&\frac12\int_0^1(1+t)dt\\ &=&\frac{3}{4},\\ g(1)&=&\int_0^1\frac1{x^3}\int_0^x\frac{t^4}{1-t}dtdx\\ &=&\int_0^1\int_0^x\frac{t^4}{x^3(1-t)}dtdx\\ &=&\int_0^1\int_t^1\frac{t^4}{x^3(1-t)}dxdt\\ &=&\frac12\int_0^1\frac{t^4}{1-t}\bigg(\frac1{t^2}-1\bigg)dt\\ &=&\frac12\int_0^1t^2(1+t)dt\\ &=&\frac{7}{24}. \end{eqnarray} So $$ \sum_{n=1}^\infty\frac1{n(n+2)(n+4)}=\frac14\bigg(\frac23-\frac{7}{24}\bigg)=\frac{11}{96}.$$

Answer 3

Just another way using partial sums.

Using Mohammad Riazi-Kermani's answer, use partial fraction decomposition to get $$a_n=\frac 1{n^3+6n^2+8n}=\frac {1}{n(n+4)(n+2)}=-\frac{1}{4 (n+2)}+\frac{1}{8 (n+4)}+\frac{1}{8 n}$$ and consider $$S_p=\sum_{n=1}^p a_n$$ Using harmonic numbers $$\sum_{n=1}^p \frac 1n=H_p\qquad \sum_{n=1}^p \frac 1{n+2}=H_{p+2}-\frac{3}{2}\qquad \sum_{n=1}^p \frac 1{n+4}=H_{p+4}-\frac{25}{12}$$

Now, use the asymptotics $$H_q=\gamma +\log \left({q}\right)+\frac{1}{2 q}-\frac{1}{12 q^2}+O\left(\frac{1}{q^3}\right)$$ Replace and continue with Taylor expansions for large $p$ to get $$S_p=\frac{11}{96}-\frac{1}{2 p^2}+O\left(\frac{1}{p^3}\right)$$ which shows the limit and also how it is approached.