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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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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

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

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

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  • $\begingroup$ This is a good hint. $\endgroup$ – Mark Bennet Oct 28 '17 at 17:51
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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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    $\begingroup$ That is a great clean answer!! +1 Not like that show off of Jack who always has to show his "ability" in complicating easy things. $\endgroup$ – Von Neumann Sep 20 '16 at 20:44
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    $\begingroup$ @FourierTransform: the benefit of complicating things is that my approach also computes $$\sum_{k\geq 1}\frac{1}{k(k+1)(k+2)(k+3)(k+4)(k+5)}$$ in two steps. It is, in fact, equivalent to telescoping. $\endgroup$ – Jack D'Aurizio Sep 20 '16 at 20:47
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    $\begingroup$ @DarioGutierrez Thanks. I just started with your identity:$\frac{1}{n(n+1)(n+2)} =\frac{1}{2}(\frac{1}{n}+\frac{1}{(n+2)})-\frac{1}{(n+1)}$, but $\frac{1}{2}(\frac{1}{n}+\frac{1}{(n+2)})-\frac{1}{(n+1)}=\frac{1}{2}\left(\frac{1}{n}-\frac{1}{(n+1)}\right)+\frac{1}{2}\left(\frac{1}{(n+2)}-\frac{1}{(n+1)}\right)$. $\endgroup$ – Olivier Oloa Sep 20 '16 at 20:51
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    $\begingroup$ @FourierTransform It's always nice to have different points of view. That was what $\texttt{Jack D'Aurizio}$ did it. Usually, I receive a lot of negative comments or/and down-votes in my own answers because I posted some weird overkill answer. What 'some' people don't understand is that I'm adding something different to the bag of answers. By the way, behind me I have a guy who behaves like a 'MSE-police officer' who down-vote, propose to close, review my $\LaTeX$, change my editions, etc... a kind of insane user. Also, I enjoy your answers too which are 'full creative'. Best Regards. $\endgroup$ – Felix Marin Sep 20 '16 at 21:43
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    $\begingroup$ @FourierTransform We should have different answers and different ideas, else, what would be the point of answering in the first place? Also, different ideas lead to learning and building of intuition, which is one reason I dislike the school system to some level. I agree with Felix, we enjoy both of your answers, and we want both. Sometimes, answers aren't meant for the OP, but for the community. $\endgroup$ – Simply Beautiful Art Sep 20 '16 at 22:04
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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? $\endgroup$ – Sanket Agrawal Jun 11 at 4:16
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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 '15 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 '15 at 14:17
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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. $\endgroup$ – Mark Bennet Oct 28 '17 at 17:51
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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 '15 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:$$\sum_{n\ge 1}\frac{1}{\sum_{j=0}^k(n+j)}=\frac{1}{k!\cdot k}.$$

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