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I have got the following series $$\sum_{k=0}^\infty \frac{1}{(2n-1)(2n+1)(2n+3)}$$

I'm trying to expand it as a telescoping series and then calculate the partial sum series but didn't succeed so far. If someone can help me expand it as a telescoping series and explain what's the technique for it it would be great.

Thanks.

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Another effective approach exploits geometric series and Euler's beta function: $$\begin{eqnarray*} \sum_{k\geq 0}\frac{1}{(2k-1)(2k+1)(2k+3)}&=&\frac{1}{8}\sum_{k\geq 0}\frac{1}{\left(k-\frac{1}{2}\right)\left(k+\frac{1}{2}\right)\left(k+\frac{3}{2}\right)}\\&=&\frac{1}{8}\sum_{k\geq 0}\frac{\Gamma\left(k-\frac{1}{2}\right)}{\Gamma\left(k+\frac{5}{2}\right)}\\&=&\frac{1}{8\,\Gamma(3)}\sum_{k\geq 0}B\left(3,k-\frac{1}{2}\right)\\&=&\frac{1}{16}\int_{0}^{1}\sum_{k\geq 0}(1-x)^2 x^{k-3/2}\,dx\\&=&\frac{1}{16}\int_{0}^{1}(1-x) x^{-3/2}\,dx\\&=&\frac{B(2,-1/2)}{16}=\color{red}{-\frac{1}{4}.}\end{eqnarray*}$$

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$$\dfrac{2n+3-(2n-1)}{(2n+3)(2n+1)(2n-1)}$$

$$=\dfrac{1}{(2n+1)(2n-1)}-\dfrac{1}{(2n+3)(2n+1)}$$

$$=F_n-F_{n+1}$$

where $$F_m=\dfrac{1}{(2m+1)(2m-1)}$$

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  • $\begingroup$ @Dr.MV, I omitted that willfully :) $\endgroup$ – lab bhattacharjee Jun 17 '16 at 17:06
  • $\begingroup$ Lab, I have restored the original version. Apology for editing. $\endgroup$ – Mark Viola Jun 17 '16 at 17:08
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HINT:

Using partial fraction expansion, we can write

$$\begin{align} \frac{1}{(2n-1)(2n+1)(2n+3)}&=\frac18\left(\frac{1}{2n-1}-\frac{2}{2n+1}+\frac{1}{2n+3}\right)\\\\ &=\frac18 \left(\frac{1}{2n-1}-\frac{1}{2n+1}\right)+\frac18 \left(\frac{1}{2n+3}-\frac{1}{2n+1}\right) \end{align}$$

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  • $\begingroup$ @georgeb George, please let me know how I can improve my answer. I really want to give you the best answer I can. -Mark $\endgroup$ – Mark Viola Jun 17 '16 at 16:58
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The technique is called partial fractions:

$$\frac{1}{(2n-1)(2n+1)(2n+3)}=\frac{A}{(2n-1)}+\frac{B}{2n+1}+\frac{C}{2n+3}$$

Now combine the denominators and equate coefficients.

$$1=A(2n+3)(2n+1)+B(2n-1)(2n+3)+C(2n-1)(2n+1)$$

$$1=A(4n^2+8n+3)+B(4n^2+4n-3)+C(4n^2-1)$$

We get:

$$4A+4B+4C=0$$

$$8A+4B=0$$

$$3A-3B-C=1$$

Substitution is often used to solve these types of equations. Note dividing the first equation leaves us with:

$$A+B+C=0$$

Hence:

$$A+B=-C$$

Substituting for $-C$ in the third equation gives:

$$3A-3B+A+B=1$$

$$4A-2B=1$$

But from the second equation we now have:

$$8A+4B=0$$

$$4A-2B=1$$

Multiply the $4A-2B=1$ equation by $2$ on both sides:

$$8A+4B=0$$

$$8A-4B=2$$

Add both equations above,

$$16A=2$$

So

$$A=\frac{1}{8}$$

Substituting this back into one of the equations that we just used elimination on we get:

$$B=-\frac{1}{4}$$

Now remember:

$$A+B=-C$$

So

$$-(A+B)=C$$

And hence,

$$C=\frac{1}{8}$$

Therefore,

$$\frac{1}{(2n-1)(2n+1)(2n+3)}=\frac{(1/8)}{(2n-1)}-\frac{2/8}{2n+1}+\frac{1/8}{2n+3}=\frac{1/8}{2n-1}-\frac{1/8}{2n+1}+\frac{1/8}{2n+3}-\frac{1/8}{2n+1}$$

I broke it down even more so it would be easier to see the cancellations.

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  • $\begingroup$ Alright i got the idea - but in this example can u show me how to isolate A B and C ? $\endgroup$ – GeorgeB Jun 17 '16 at 16:52
  • $\begingroup$ You've got to know what form to use so you can equate coefficients see en.m.wikibooks.org/wiki/Calculus/Integration_techniques/… $\endgroup$ – Ahmed S. Attaalla Jun 17 '16 at 16:52
  • $\begingroup$ Yes I will @GeorgeB $\endgroup$ – Ahmed S. Attaalla Jun 17 '16 at 16:53
  • $\begingroup$ Thanks, I'm familiar with this technique from my calculus course , but still found it hard to implement here. $\endgroup$ – GeorgeB Jun 17 '16 at 16:54
  • $\begingroup$ How is this approach different from the one I posted? $\endgroup$ – Mark Viola Jun 17 '16 at 17:01

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