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Problem: Find the sum to $n$ terms of \begin{eqnarray*} \frac{1}{1\cdot 2\cdot 3} + \frac{3}{2\cdot 3\cdot 4} + \frac{5}{3\cdot 4\cdot 5} + \frac{7}{4\cdot 5\cdot 6}+\cdots \\ \end{eqnarray*} Answer: The way I see it, the problem is asking me to find this series: \begin{eqnarray*} S_n &=& \sum_{i=1}^{n} {a_i} \\ \text{with } a_i &=& \frac{2i-1}{i(i+1)(i+2)} \\ \end{eqnarray*} We have: \begin{eqnarray*} S_n &=& S_{n-1} + a_n \\ S_n &=& S_{n-1} + \frac{2n-1}{n(n+1)(n+2)} \\ \end{eqnarray*} I am tempted to apply the technique of partial fractions but I believe there is no closed formula for a series of the of the form:

\begin{eqnarray*} \sum_{i=1}^{n} \frac{1}{i+k} \\ \end{eqnarray*} where $k$ is a fixed constant. Therefore I am stuck. I am hoping that somebody can help me.

Thanks Bob

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  • $\begingroup$ Maple says that $S_n = \frac{1}{2(n+1)}-\frac{5}{2(n+2)}+\frac34$. $\endgroup$ – Michael Galuza Jun 27 '15 at 20:00
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There is more simple way (for me). You have $$a_n=\frac{2n-1}{n(n+1)(n+2)}=-\frac52\frac{1}{n+2} + \frac{3}{n+1} - \frac{1}{2n};$$ hence $$S_N = \sum_{n=1}^N a_n = -\frac52\left(H_{N+2}-1-\frac12\right) + 3(H_{N+1}-1) - \frac12 H_N,$$ where $H_N$ is $N$-th harmonic number. Simplify it: $$S_N = -\frac52\left(H_N + \frac{1}{N+1} + \frac{1}{N+2}-\frac32\right) + 3\left(H_N+\frac{1}{N+1}-1\right) - \frac12H_N=\\= \frac34 + \frac{1}{2(N+1)}-\frac{5}{2(N+2)}.$$

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The good old way: compute a few terms and find a pattern.

$$\frac2{12},\frac7{24},\frac{15}{40},\frac{26}{60},\frac{40}{84}\cdots$$

By finite differences, we find that the numerators are $\dfrac{n(3n+1)}2$ and the denominators $2(n+1)(n+2)$.


Check:

$$S_1=\frac2{12}$$and $$S_n-S_{n-1}=\frac{n(3n-1)}{4(n+1)(n+2)}-\frac{(n-1)(3n-4)}{4n(n+1)}=\frac{2n-1}{n(n+1)(n+2)}=a_n.$$

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Setting

$$\frac{2n-1}{n(n+1)(n+2)}=\frac{An+B}{n(n+1)}-\frac{A(n+1)+B}{(n+1)(n+2)}$$ gives us $A=2,B=-\frac 12$, i.e. $$\frac{2n-1}{n(n+1)(n+2)}=\frac{2n-\frac 12}{n(n+1)}-\frac{2(n+1)-\frac 12}{(n+1)(n+2)}.$$

Hence, we have $$\begin{align}\sum_{i=1}^{n}\frac{2i-1}{i(i+1)(i+2)}&=\sum_{i=1}^{n}\left(\frac{2i-\frac 12}{i(i+1)}-\frac{2(i+1)-\frac 12}{(i+1)(i+2)}\right)\\&=\frac{2\cdot 1-\frac 12}{1\cdot (1+1)}-\frac{2(n+1)-\frac 12}{(n+1)(n+2)}\\&=\color{red}{\frac 34-\frac{4n+3}{2(n+1)(n+2)}}\end{align}$$

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You can write it as $\sum_{k\ge0}\frac{2k+1}{(k+3)_3}=\sum_{k\ge0}(2k+1)(k)_{-3}$ and now it seems easy to solve by summation by parts:

$$\sum (2k+1)(k)_{-3}\delta k=(2k+1)\frac{(k)_{-2}}{-2}+\sum(k+1)_{-2}=\frac{2k+1}{-2(k+2)_2}-\frac{1}{k+2}=\frac{4k+3}{-2(k+2)_2}$$

And taking limits we have that the series converges to $\sum_{k\ge 0}\frac{2k+1}{(k+3)_3}=\frac{3}{4}$

The partial sum will be

$$\sum_{k=0}^{n}\frac{2k+1}{(k+3)_3}=\frac{4n+3}{-2(n+2)_2}+\frac{3}{4}$$

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  • $\begingroup$ (+1) Very neat. It is a good thing to spread the use of Pochhammer symbols and summation by parts. $\endgroup$ – Jack D'Aurizio Jun 27 '15 at 21:12
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    $\begingroup$ Thnk you @JackD'Aurizio... anyway I must point here that this isnt, strictly talking, Pochhammer symbol because Pochhammer used it to denote rising factorials and in this case Im denoting falling factorials. I may be use the notation of Knuth to avoid confusion... but this notation, still with this "lack" of universality, is faster to write. And this is the reason why I used it :) $\endgroup$ – Masacroso Jun 27 '15 at 21:23
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see that $a_n = (2n-1)/((n*(n+1)*(n+2)=-1/2*(1/n)+3*(1/(n+1))-5/2*(1/(n+2)) = 1/2* (-1/n+1/(n+1)+5*(1/(n+1)-1/(n+2))$, so $S_n = 1/2*(-1+5+1/(n+1)-5/(n+2)) = 1/2*(4n^2+8n+5)/((n+1)*(n+2))$.

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