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Find the sum of order:

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

There is how they count it in book:

$$s_{n} = \left(\frac{3}{10}-\frac{3}{2}\right)+\left(\frac{3}{14}-\frac{1}{2}\right)+\left(\frac{1}{6}-\frac{3}{10}\right)+\left(\frac{3}{22}-\frac{3}{14}\right)+...+\left(\frac{3}{4n-2}-\frac{3}{4n-10}\right)+\left(\frac{3}{4n+2}-\frac{3}{4n-6}\right)+\left(\frac{3}{4n+6}-\frac{3}{4n-2}\right)$$ $$s_{n} = \frac{-3}{2}+\frac{-1}{2}+\frac{3}{4n+2}+\frac{3}{4n+6}$$ $$s = \lim_{n->∞}s_{n} = \lim_{n->∞}\left [\frac{-3}2 - \frac12 + \frac3{4n+2} + \frac3{4n+6}\right ] = -2$$

I understand how to solve the lim, I just dont understand, how to get those last items from order, I mean this items:

$$\left(\frac{3}{4n-2}-\frac{3}{4n-10}\right)+\left(\frac{3}{4n+2}-\frac{3}{4n-6}\right)+\left(\frac{3}{4n+6}-\frac{3}{4n-2}\right)$$

UPDATE

Now I understand how to get those "last" items. But I'm confused now, why are they even in the sum inside of lim of $s_n$? If I would keep counting next items, they would get canceled. For example, there is $\frac3{4n+2}$ in $s_n$, if I count n+1 item, this would get canceled. So why do we count them in $s_n$ if only first two fractals {$\frac{-3}2, \frac12 $} couldn't be canceled (if not thinking of negative n). Could anyone explain please?

Buksy
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  • I have used this editor: http://www.codecogs.com/latex/eqneditor.php for equations, sorry ... how to fix them? – Buksy Mar 05 '13 at 10:28
  • Your $\LaTeX$ was fine, you only didn't take any environments, so Mathjax didn't know what to render – Dominic Michaelis Mar 05 '13 at 10:29
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    @Buksy: You just needed to wrap everything in dollar signs, $...math stuff...$, or double dollar signs to use "display" mode $$...math stuff...$$. – Zev Chonoles Mar 05 '13 at 10:29
  • You can find some good starting points on how to format mathematics on the site here. This AMS reference is very useful. If you need to format more advanced things, there are many excellent references on LaTeX on the internet, including StackExchange's own TeX.SE site. – Zev Chonoles Mar 05 '13 at 10:29
  • There should be a $+ \cdots$ to indicate that the sequence continues. Those are not the "last" items; it is an attempt to show which terms cancel when the sequence telescopes down. – Peter Phipps Mar 05 '13 at 12:04
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    There is a common factor of $\frac{3}{2}$ in all the terms. Take it out: typing gets easier. More importantly, it will make the structure clearer. – André Nicolas Mar 05 '13 at 18:45
  • Thank you all, could you please take a look at my update? – Buksy Mar 07 '13 at 12:42
  • Should I create new question for my update to this question or is it ok keep it here? – Buksy Mar 08 '13 at 21:43

3 Answers3

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I hope i didn't missunterstood your question, this sum is a so called telescoping sum, the other terms are canceled so you only have those left which are mentioned.

Those are

\begin{align*} \frac{3}{4n-10}&= \frac{3}{4(n-1)-6}\\ \frac{3}{4n-6}&= \frac{3}{4(n-1)-2} \\ \end{align*} And $$\frac{3}{4n-2}$$ is canceled from the last paranthesis.

Dominic Michaelis
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The 3 terms you have mentioned above saying you didnot understand are the (n-2),(n-1) and nth terms of that series.

So if you just substitute 'n' with 'n-2' you get : $$\frac{3}{4n-2} - \frac{3}{4n-10}$$

if you substitute 'n' with 'n-1' you get : $$\frac{3}{4n+2} - \frac{3}{4n-6}$$

Buksy
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lsp
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For the update: If you look at the denominators, the first fraction goes $5,7,9,11,\ldots$ The second fraction goes $1,3,5,7,9\ldots$ As they enter with opposite signs, all the matching ones can be canceled with one from the other set. We are left with just $1$ and $3$. Because these are in the deominators, that is what becomes $\frac {-\frac 32}1 + \frac {-\frac 32}3=-2$. This is commonly true for telescoping sums. The surviving terms are the first few, which are not canceled. All the rest vanish through cancellation.

Ross Millikan
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