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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.

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10 Answers 10

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Hint

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

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

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    $\begingroup$ This is a good hint. $\endgroup$ Oct 28, 2017 at 17:51
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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?

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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 $$ \sum_{n=1}^N\frac{1}{n(n+1)(n+2)} =\frac{1}{2}-\frac{1}{2(N+1)}+\frac{1}{2(N+2)}-\frac{1}{4}. $$

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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: $$ \sum_{n\geq 1}\frac{1}{n(n+1)\cdots(n+N)}=\color{red}{\frac{1}{N\cdot N!}}.$$

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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

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

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    $\begingroup$ This is the best form to use - simple and straightforward. $\endgroup$ Oct 28, 2017 at 17:51
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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}

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  • $\begingroup$ I know this method and answer is correct, but how do I justify interchanging the summation and integration sign? $\endgroup$ Jun 11, 2019 at 4:16
  • $\begingroup$ using the dominating convergence theorem $\endgroup$
    – inoxb911
    Apr 7, 2022 at 10:15
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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.

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  • $\begingroup$ 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? $\endgroup$
    – Clement C.
    Mar 26, 2015 at 14:12
  • $\begingroup$ I suppose, since you also have to take the limit to get $0\log{0} "=" 0$. I'll put a note. $\endgroup$
    – Chappers
    Mar 26, 2015 at 14:17
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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.)

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  • $\begingroup$ Interesting approach... Different from everything I learned.... Thanks! $\endgroup$
    – Frank
    Mar 26, 2015 at 14:09
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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:$$\sum_{n\ge 1}\frac{1}{\prod_{j=0}^k(n+j)}=\frac{1}{k!\cdot k}.$$

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Let me add a more general answer.

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 $$ \eqalign{ & {1 \over {n\left( {n + 1} \right)\left( {n + 2} \right)}} = {1 \over {n^{\,\overline {\,3\,} } }} = \left( {n - 1} \right)^{\,\underline {\, - 3\,} } \cr & \sum\nolimits_n {{1 \over {n\left( {n + 1} \right)\left( {n + 2} \right)}}} = \sum\nolimits_n {\left( {n - 1} \right)^{\,\underline {\, - 3\,} } } = \cr & = - {1 \over 2}\left( {n - 1} \right)^{\,\underline {\, - 2\,} } + c = - {1 \over {2n^{\,\overline {\,2\,} } }} + c = - {1 \over {2n\left( {n + 1} \right)}} + c \cr & \sum\limits_{n = 1}^m {{1 \over {n\left( {n + 1} \right)\left( {n + 2} \right)}}} = \sum\nolimits_{n = 1}^{m + 1} {\left( {n - 1} \right)^{\,\underline {\, - 3\,} } } = \cr & = - {1 \over 2}\left( {{1 \over {\left( {m + 1} \right)\left( {m + 2} \right)}} - {1 \over {1 \cdot 2}}} \right) = {1 \over 2}\left( {{1 \over 2} - {1 \over {\left( {m + 1} \right)\left( {m + 2} \right)}}} \right) \cr & \sum\limits_{n = 1}^\infty {{1 \over {n\left( {n + 1} \right)\left( {n + 2} \right)}}} = {1 \over 4} \cr} $$

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