How to find last items of order to get its sum 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?

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 : $$\frac{3}{4n+2} - \frac{3}{4n-6}$$
