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Show $$\sum_{n=1}^\infty {\frac{1}{(n+1)(n+2)(n+3)}}$$ is convergent using partial fractions and find limit.

I know how to do this by integral test and surely if you just multiply out the brackets you can use comparison test? Even ratio test could work? Anyway in this question we have to use the following hint:

This is a so-called telescopic series. Find an explicit representation for the partial sums and prove this.

I don't know how to do this, if you make into partial fractions you have three fractions none of which converge..

Partial fractions gives: $\frac{1}{2(n+1)}+\frac{1}{2(n+3)}-\frac{1}{n+2}$

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    $\begingroup$ Please write down what you obtained after doing partial fraction decomposition. $\endgroup$ – user258700 Dec 8 '15 at 16:51
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    $\begingroup$ Why would the limit be zero when you sum only positive values? $\endgroup$ – Hetebrij Dec 8 '15 at 16:51
  • $\begingroup$ I've added in result of partial fractions $\endgroup$ – babylon Dec 8 '15 at 16:55
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    $\begingroup$ It is true that the convergence is obvious. (But Ratio Test is inconclusive.) However, you are being asked to find an explicit number for the sum, Yes, if you break it up into $3$ series, none will converge. But if you combine terms suitably, magic will happen. $\endgroup$ – André Nicolas Dec 8 '15 at 16:56
  • $\begingroup$ I like magic. ${}{}$ $\endgroup$ – copper.hat Dec 8 '15 at 17:03
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Hint: $$\frac{1}{(n+1)(n+2)(n+3)}=\frac{1}{2}\left(\frac{1}{(n+1)(n+2)}-\frac{1}{(n+2)(n+3)}\right).$$ Or, using your partial fractions:

$$\frac{1}{(n+1)(n+2)(n+3)} = \left(\frac{1}{2(n+1)}-\frac1{2(n+2)}\right)-\left(\frac{1}{2(n+2)}-\frac1{2(n+3)}\right)$$

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You have ${1 \over (n+1)(n+2)(n+3)} = {1 \over 2}({1 \over n+1} - {1 \over n+2}) + {1 \over 2}({1 \over n+3} - {1 \over n+2})$. The latter two are telescoping sums.

Hence the sum is

${1 \over 2}{1 \over 1+1} - {1 \over 2}{1 \over 1+2} = {1 \over 12}$.

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BIG HINT:

$$\sum_{n=1}^\infty {\frac{1}{(n+1)(n+2)(n+3)}}=$$ $$\lim_{m\to\infty}\sum_{n=1}^m{\frac{1}{(n+1)(n+2)(n+3)}}=$$ $$\lim_{m\to\infty}\sum_{n=1}^m\frac{1}{2}\left(\frac{1}{(n+1)(n+2)}-\frac{1}{(n+2)(n+3)}\right)=$$ $$\frac{1}{2}\lim_{m\to\infty}\sum_{n=1}^m\left(\frac{1}{(n+1)(n+2)}-\frac{1}{(n+2)(n+3)}\right)=$$ $$\frac{1}{2}\lim_{m\to\infty}\left(\frac{1}{6}-\frac{1}{m^2+5m+6}\right)=$$ $$\frac{1}{2}\left(\lim_{m\to\infty}\frac{1}{6}-\lim_{m\to\infty}\frac{1}{m^2+5m+6}\right)=$$ $$\frac{1}{2}\left(\frac{1}{6}-\lim_{m\to\infty}\frac{1}{m^2+5m+6}\right)=$$

$$\frac{1}{2}\left(\frac{1}{6}-\frac{1}{\lim_{m\to\infty}\left(m^2+5m+6\right)}\right)$$

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Notice, $$\sum_{n=1}^{\infty}\frac{1}{(n+1)(n+2)(n+3)}$$ $$=\frac{1}{2}\sum_{n=1}^{\infty}\left(\frac{1}{(n+1)(n+2)}-\frac{1}{(n+2)(n+3)}\right)$$ $$=\frac{1}{2}\lim_{n\to\infty}\left[\left(\frac{1}{2\cdot 3}-\frac{1}{3\cdot 4}\right)+\left(\frac{1}{3\cdot 4}-\frac{1}{4\cdot 5}\right)+\ldots+\left(\frac{1}{(n+1)(n+2)}-\frac{1}{(n+2)(n+3)}\right)\right]$$ $$=\frac{1}{2}\lim_{n\to\infty}\left[\frac{1}{2\cdot 3}-\frac{1}{(n+2)(n+3)}\right]$$ $$=\frac{1}{2}\left[\frac{1}{6}-0\right]=\color{red}{\frac{1}{12}}$$

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$$\sum_{n=1}^\infty {\frac{1}{(n+1)(n+2)(n+3)}}$$ $$=\frac12\sum_{n=1}^\infty \left(\frac1{n+1}-\frac2{n+2}+\frac1{n+3}\right)$$ $$=\frac12\sum_{n=1}^\infty \left(\frac1{n+1}-\frac1{n+2}\right)-\frac12\sum_{n=1}^\infty \left(\frac1{n+2}-\frac1{n+3}\right)$$ $$=\frac12\left(\frac1{2}-\lim_{n\to\infty}\frac1{n+2}\right)-\frac12\left(\frac1{3}-\lim_{n\to\infty}\frac1{n+3}\right)$$ $$=\frac12\cdot\frac12-\frac12\cdot\frac13=\frac{1}{12}$$

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  • $\begingroup$ I think the last manipulation should be done with partial sums, because, since the armonic series is divergent, you might not be allowed to use infinite times the associative property in $\sum_{n=1}^\infty \left(\frac{1}{n+1}-\frac{1}{n+2}\right)$ $\endgroup$ – user228113 Dec 8 '15 at 17:13
  • $\begingroup$ This $\sum_{n=1}^\infty \left(\frac1{n+1}-\frac1{n+2}\right)$ alone converges. $\endgroup$ – Kay K. Dec 8 '15 at 17:16
  • $\begingroup$ I agree. And it converges absolutely. And so does $$\sum_{k=0}^\infty ((-1)^k+(-1)^{k+1})=0$$ but, if you do the same manipulation you did to get $\frac12$ in your proof, you'll get $$(-1)^0+\sum_{k=1}^{\infty} 2(-1)^k$$ which is not convergent. $\endgroup$ – user228113 Dec 8 '15 at 17:20
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    $\begingroup$ $\sum_{n=1}^\infty \left(\frac1{n+1}-\frac1{n+2}\right)=\frac12$ and $\sum_{n=1}^\infty \left(\frac1{n+2}-\frac1{n+3}\right)=\frac13$, which is what the last line is saying. $\endgroup$ – Kay K. Dec 8 '15 at 17:25
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    $\begingroup$ Now I think I get what you were trying to tell me. Yes I skipped the proof. I agree it's good to include it. I'll add it. Thanks. $\endgroup$ – Kay K. Dec 8 '15 at 17:32

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