# Find the sum of the infinite series [closed]

Find the sum of the infinite series: \begin{align*}\frac{1}{2.3.4}+\frac{1}{4.5.6} +\frac{1}{6.7.8}+\frac{1}{8.9.10} + \ldots &= \sum_{n=1}^{\infty}\frac{1}{2n(2n+1)(2n+2))} \\ &= \frac{1}{4}\sum_{1}^{\infty}\frac{1}{n(n+1)(2n+1)} \\ &=\sum_{1}^{\infty}\frac{1}{4n}-\sum_{1}^{\infty}\frac{1}{2n+1}+\sum_{1}^{\infty}\frac{1}{2(2n+2)} \\ &= \frac14+\frac14+\frac16+\ldots 1-\frac12-\frac14-\frac16-\ldots -1+\frac12-\frac13+\frac14 \ldots \\ &=1+\frac14-\frac12-\ln2 \\ &=\frac{1}{4}[3-4\ln2] \end{align*}

• $$\frac{1}{n(n+1)(n+2)}=\frac{1/2}{n}-\frac{1}{n+1}+\frac{1/2}{n+2}$$
– BAI
Aug 6, 2017 at 14:01
• Add to BAI, I think use $\frac{1}{n+1}=\frac{1/2}{n+1}+\frac{1/2}{n+1}$ .... you will get a telescopic sum, I guess. Aug 6, 2017 at 14:07
• I don't understand why it is still written "on hold"? Aug 7, 2017 at 17:54

$$\sum_{n=1}^{\infty}\frac{1}{2n(2n+1)(2n+2))}=\frac{1}{4}\sum_{1}^{\infty}\frac{1}{n(n+1)(2n+1)}=\frac{1}{4}[3-4\ln2]$$

• How have you reached at the last equality? Aug 6, 2017 at 15:01
• Prudnikov, Brychkov, Marichev, Integrals and Series Volume 1 Elementary Functions, 1986, pg 666. Aug 6, 2017 at 15:31
• It's done thanks alot. Aug 7, 2017 at 3:29
• I checked it's ans is $\frac34-ln2$ Aug 7, 2017 at 17:11