# Test the condition for convergence of \sum_{n=1}^\infty\frac{1}{n(n+1)(n+2)} and find the sum if

Sandra Allison 2022-01-18 Answered
Test the condition for convergence of
$\sum _{n=1}^{\mathrm{\infty }}\frac{1}{n\left(n+1\right)\left(n+2\right)}$
and find the sum if it exists.
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David Clayton
Karen Robbins

There is an alternate method and is as follows.
Notice that
$\frac{1}{n\left(n+1\right)\left(n+2\right)}=\frac{\left(n-1\right)!}{\left(n+2\right)!}=\frac{12}{B}\left(n,3\right)$
where $B\left(x,y\right)$ is the Beta function. Using an integral form of the Beta function the summation becomes
$S=\sum _{n=1}^{\mathrm{\infty }}\frac{1}{n\left(n+1\right)\left(n+2\right\}}$
$=\frac{1}{2}{\int }_{0}^{1}\left(\sum _{n=1}^{\mathrm{\infty }}{x}^{n-1}\right)\left(1-x{\right)}^{2}dx$
$=\frac{1}{2}{\int }_{0}^{1}\frac{\left(1-x{\right)}^{2}}{1-x}dx=\frac{1}{2}{\int }_{0}^{1}\left(1-x\right)dx$
$=\frac{1}{4}$
This leads to the known result
$\sum _{n=1}^{\mathrm{\infty }}\frac{1}{n\left(n+1\right)\left(n+2\right)}=\frac{1}{4}$

###### Not exactly what you’re looking for?
alenahelenash
Alternatively, take $\frac{1}{1-x}=\sum _{n=1}^{\mathrm{\infty }}{x}^{n-1}$ and integrate three times with lower limit 0, giving $-\mathrm{log}\left(1-x\right)=\sum _{n=1}^{\mathrm{\infty }}\frac{{x}^{n}}{n}$ $x+\left(1-x\right)\mathrm{log}\left(1-x\right)=\sum _{n=1}^{\mathrm{\infty }}\frac{{x}^{n}}{n\left(n+1\right)}$ $\frac{3}{4}{x}^{2}-\frac{1}{2}x-\frac{1}{2}\left(1-x{\right)}^{2}\mathrm{log}\left(1-x\right)=\sum _{n=1}^{\mathrm{\infty }}\frac{{x}^{n}}{n\left(n+1\right)\left(n+2\right)}$