# Find the sum of the series $\sum \frac{1}{n(n+1)(n+2)}$

I got this question in my maths paper

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

I managed to show that the series converges but I was unable to find the sum. Any help/hint will go a long way. Thank you.

• You could calculate the sum up to $N$ for $N=1,2,3,\dots$, guess a pattern to the answers, then prove the pattern persists by induction, then take the limit (but the answer by N. S. looks better). Nov 10, 2013 at 5:48
• Comments are not for extended discussion; this conversation has been moved to chat. Sep 21, 2016 at 11:48
• See also this post: $\sum _{n=1}^{\infty} \frac{1}{n (n+1) (n+2)}$ Understand the representation. Some answers there provide nice visual explanation how the terms in the telescoping series are cancelled. Oct 28, 2016 at 10:21

Hint

$$\frac{2}{n(n+2)}=\frac{1}{n}-\frac{1}{n+2}$$

Now multiply both sides by $\frac{1}{n+1}$.

• This is a good hint. Oct 28, 2017 at 17:51

Using Partial Fraction Decomposition, $$\frac1{n(n+1)(n+2)}=\frac An+\frac B{n+1}+\frac C{n+2}$$

$$\implies 1=A(n+1)(n+2)+Bn(n+2)+Cn(n+1)$$ $$\implies 1=n^2(A+B+C)+n(3A+2B+C)+2A$$

Comparing the coefficients of the different powers (namely, $0,1,2$) of $n,$ we get $A=\frac12,B=-1,C=\frac12$

$$\implies\frac1{n(n+1)(n+2)}=\frac12\cdot\frac1n-\frac1{n+1}+\frac12\cdot\frac1{n+2}$$ $$=-\frac12\left(\underbrace{\frac1{n+1}-\frac1n}\right)+\frac12\left(\underbrace{\frac1{n+2}-\frac1{n+1}}\right)$$

Can you recognize the two Telescoping series?

Hint. You may write $$\frac{1}{n(n+1)(n+2)} =\frac{1}{2}\left(\frac{1}{n}-\frac{1}{(n+1)}\right)+\frac{1}{2}\left(\frac{1}{(n+2)}-\frac{1}{(n+1)}\right)$$ giving two telescoping sums $$\sum_{n=1}^N\frac{1}{n(n+1)(n+2)} =\frac{1}{2}-\frac{1}{2(N+1)}+\frac{1}{2(N+2)}-\frac{1}{4}.$$

Using Euler's beta function, $$S=\sum_{k\geq 1}\frac{(k-1)!}{(k+2)!}=\sum_{k\geq 1}\frac{\Gamma(k)}{\Gamma(k+3)}=\frac{1}{2}\sum_{k\geq 1}B(k,3)$$ hence: $$S = \frac{1}{2}\int_{0}^{1}(1-x)^2\sum_{k\geq 1}x^{k-1}\,dx=\frac{1}{2}\int_{0}^{1}(1-x)\,dx = \color{red}{\frac{1}{4}}.$$ A straightforward generalization of this approach gives the identity: $$\sum_{n\geq 1}\frac{1}{n(n+1)\cdots(n+N)}=\color{red}{\frac{1}{N\cdot N!}}.$$

Writing

$$\frac{1}{n(n+1)(n+2)} = \frac{1}{n+1}\left(\frac{1}{2n} - \frac{1}{2(n+2)}\right) = \frac{1}{2n(n+1)} - \frac{1}{2(n+1)(n+2)},$$

we see that the sum telescopes to

$$\frac{1}{2(1)(2)} = \frac{1}{4}.$$

• This is the best form to use - simple and straightforward. Oct 28, 2017 at 17:51

There is an alternate method and is as follows.

Notice that $$\frac{1}{n(n+1)(n+2)} = \frac{(n-1)!}{(n+2)!} = \frac{1}{2!} \, B(n,3)$$ where $B(x,y)$ is the Beta function. Using an integral form of the Beta function the summation becomes \begin{align} S &= \sum_{n=1}^{\infty} \frac{1}{n \, (n+1) \, (n+2)} \\ &= \frac{1}{2} \, \int_{0}^{1} \left( \sum_{n=1}^{\infty} x^{n-1} \right) \, (1-x)^{2} \, dx \\ &= \frac{1}{2} \, \int_{0}^{1} \frac{(1-x)^{2}}{1-x} \, dx = \frac{1}{2} \, \int_{0}^{1} (1-x) \, dx \\ &= \frac{1}{4} \end{align}

This leads to the known result \begin{align} \sum_{n=1}^{\infty} \frac{1}{n \, (n+1) \, (n+2)} = \frac{1}{4}. \end{align}

• I know this method and answer is correct, but how do I justify interchanging the summation and integration sign? Jun 11, 2019 at 4:16
• using the dominating convergence theorem Apr 7, 2022 at 10:15

Alternatively, take $$\frac{1}{1-x} = \sum_{n=1}^{\infty} x^{n-1},$$ and integrate three times with lower limit $0$, giving \begin{align*} -\log{(1-x)} &= \sum_{n=1}^{\infty} \frac{x^n}{n} \\ x + (1-x)\log{(1-x)} &= \sum_{n=1}^{\infty} \frac{x^n}{n(n+1)} \\ \frac{3}{4}x^2 - \frac{1}{2}x - \frac{1}{2} (1-x)^2 \log{(1-x)} &= \sum_{n=1}^{\infty} \frac{x^n}{n(n+1)(n+2)}, \end{align*} and (as @Clement C reminds me) we then apply Abel's theorem to take the limit as $x \to 1$, which gives $1/4$ as the answer.

• Isn't there a bit more to the argument, then? The original power series has radius of convergence $1$, so the equality and all integration theorems only apply on $(-1,1)$. To apply the conclusion to $x=1$ at the end (outside the open disc of convergence of the original series, and where the final LHS is only defined by continuity), don't you need Abel's theorem? Mar 26, 2015 at 14:12
• I suppose, since you also have to take the limit to get $0\log{0} "=" 0$. I'll put a note. Mar 26, 2015 at 14:17

You are on the right track. Fix $N \geq 1$. Then \begin{align} \sum_{n=1}^{N} \frac{1}{n \cdot (n+1) \cdot (n+2)} &= \sum_{n=1}^N \left(\frac{1}{2n}-\frac{1}{n+1}+\frac{1}{2n+4}\right) = \frac{1}{2}\sum_{n=1}^N \frac{1}{n}-\sum_{n=1}^{N}\frac{1}{n+1}+\frac{1}{2}\sum_{n=1}^{N} \frac{1}{n+2}\\ &= \frac{1}{2}\sum_{n=1}^N \frac{1}{n}-\sum_{n=2}^{N+1}\frac{1}{n}+\frac{1}{2}\sum_{n=3}^{N+2} \frac{1}{n} \\ &= \frac{1}{2}\left(1+\frac{1}{2}\right) + \frac{1}{2}\sum_{n=3}^N \frac{1}{n}-\left(\frac{1}{2}+\frac{1}{N+1}\right)-\sum_{n=3}^{N}\frac{1}{n}+\\ &\quad \left(\frac{1}{N+1}+\frac{1}{N+2}\right)+\frac{1}{2}\sum_{n=3}^{N} \frac{1}{n} \\ \end{align}

Can you continue from there? (the partial sums cancel out, and you only have a few remaining terms. Taking the limit $N\to\infty$ will give you the limit.)

• Interesting approach... Different from everything I learned.... Thanks! Mar 26, 2015 at 14:09

Here's another way to do it:$$\sum_{n\ge 1}\frac{1}{n(n+1)(n+2)}=\frac{1}{2}\sum_{n\ge 1}\bigg(\frac{1}{n(n+1)}-\frac{1}{(n+1)(n+2)}\bigg)=\frac{1}{2}\cdot\frac{1}{1\cdot 2}=\frac{1}{4}.$$One advantage is an easy generalisation to a problem for which the partial fractions decomposition would get thorny:$$\sum_{n\ge 1}\frac{1}{\prod_{j=0}^k(n+j)}=\frac{1}{k!\cdot k}.$$

Note that $${1 \over {\left( {x + 1} \right)\left( {x + 2} \right) \cdots \left( {x + m} \right)}} = {1 \over {\left( {x + 1} \right)^{\,\overline {\,m\,} } }} = {{\Gamma \left( {x + 1} \right)} \over {\Gamma \left( {x + 1 + m} \right)}} = x^{\,\underline {\, - m\,} }$$ where $$x^{\,\underline {\,k\,} } ,\quad x^{\,\overline {\,k\,} }$$ represent respectively the Falling and Rising Factorial.
In virtue of the definition through the Gamma function, the above is valid for $$x,m \in \mathbb C$$.
Then we apply the Indefinite Sum concept, by which \eqalign{ & \Delta _{\,x} \;x^{\,\underline {\,q\,} } = \left( {x + 1} \right)^{\,\underline {\,q\,} } - x^{\,\underline {\,q\,} } = qx^{\,\underline {\,q - 1\,} } \quad \Rightarrow \cr & \Rightarrow \quad \Delta _{\,x} ^{\left( { - 1} \right)} \;x^{\,\underline {\,q\,} } = \sum\nolimits_x {x^{\,\underline {\,q\,} } } = \left\{ {\matrix{ {{1 \over {q + 1}}\;x^{\,\underline {\,q + 1\,} } + c} & { - 1 \ne q} \cr {\psi (x + 1) + c} & { - 1 = q} \cr } } \right. \cr} which is to recall that it is valid for $$x,q,c \in \mathbb C$$.
So in particular, for $$q=-m$$ and $$m \ne 1$$, the Indefinite sum becomes $$\sum\nolimits_x {x^{\,\underline {\, - m\,} } } \quad \left| {\,1 \ne m} \right. = {1 \over { - m + 1}}x^{\,\underline {\, - m + 1\,} } + c$$ which means that for the definite sum we get \eqalign{ & \sum\limits_{x = a}^b {x^{\,\underline {\, - m\,} } } = \sum\nolimits_{x = a}^{b + 1} {x^{\,\underline {\, - m\,} } } \quad \left| \matrix{ \,m,a,b \in C \hfill \cr \;1 \ne m \hfill \cr} \right. = \cr & = {1 \over { - m + 1}}\left( {\left( {b + 1} \right)^{\,\underline {\, - m + 1\,} } - a^{\,\underline {\, - m + 1\,} } } \right) = \cr & = {1 \over {m - 1}}\left( {a^{\,\underline {\, - \left( {m - 1} \right)\,} } - \left( {b + 1} \right)^{\,\underline {\, - \left( {m - 1} \right)\,} } } \right) = \cr & = {1 \over {m - 1}}\left( {{1 \over {\left( {a + 1} \right)^{\,\overline {\,m - 1\,} } }} - {1 \over {\left( {b + 2} \right)^{\,\overline {\,m - 1\,} } }}} \right) \cr} and taking the limit $$b \to \infty$$ $$\sum\limits_{x = a}^\infty {x^{\,\underline {\, - m\,} } } \quad \left| {\,\;1 < m \in R} \right. = {1 \over {\left( {m - 1} \right)m!}}$$
Finally, in your particular case, the summand, the indefinite, definite and infinite sums are \eqalign{ & {1 \over {n\left( {n + 1} \right)\left( {n + 2} \right)}} = {1 \over {n^{\,\overline {\,3\,} } }} = \left( {n - 1} \right)^{\,\underline {\, - 3\,} } \cr & \sum\nolimits_n {{1 \over {n\left( {n + 1} \right)\left( {n + 2} \right)}}} = \sum\nolimits_n {\left( {n - 1} \right)^{\,\underline {\, - 3\,} } } = \cr & = - {1 \over 2}\left( {n - 1} \right)^{\,\underline {\, - 2\,} } + c = - {1 \over {2n^{\,\overline {\,2\,} } }} + c = - {1 \over {2n\left( {n + 1} \right)}} + c \cr & \sum\limits_{n = 1}^m {{1 \over {n\left( {n + 1} \right)\left( {n + 2} \right)}}} = \sum\nolimits_{n = 1}^{m + 1} {\left( {n - 1} \right)^{\,\underline {\, - 3\,} } } = \cr & = - {1 \over 2}\left( {{1 \over {\left( {m + 1} \right)\left( {m + 2} \right)}} - {1 \over {1 \cdot 2}}} \right) = {1 \over 2}\left( {{1 \over 2} - {1 \over {\left( {m + 1} \right)\left( {m + 2} \right)}}} \right) \cr & \sum\limits_{n = 1}^\infty {{1 \over {n\left( {n + 1} \right)\left( {n + 2} \right)}}} = {1 \over 4} \cr}