Test the convergence of the series
$\displaystyle \frac{1}{1.2.3}+\frac{3}{2.3.4}+\frac{5}{3.4.5}+.. .$
Split that into two series (both of them converge because of direct comparison test) and then:
$\displaystyle \frac{1}{(n+1)(n+2)}=\frac{n!}{(n+2)!}=\frac{\Gamm a (n+1)\Gamma (2)}{\Gamma (n+3)}=\beta (n+1,2).$
Hence, turn that into a beta integral and switch the integral by the series. Use geometric series and solve the remaining integral. Do the same with the other.
In general for $\displaystyle n\ge2,$ $\displaystyle \sum\limits_{k=0}^{\infty }{\frac{1}{(k+1)(k+2)\cdots (k+n)}}=\frac{1}{(n-1)(n-1)!},$ which can be proved in the same fashion when using Beta function.
There's also a very nice pattern if you decompose the series, i.e.
$\displaystyle \frac{1}{n(n+1) }= \frac{1}{n}-\frac{1}{n+1} $
$\displaystyle \frac{1}{n(n+1) (n+2)}= \frac{1}{2}\left(\frac{1}{n}-\frac{2}{n+1} + \frac{1}{n+2}\right) $
$\displaystyle \frac{1}{n(n+1) (n+2)(n+3)}= \frac{1}{6}\left(\frac{1}{n}-\frac{3}{n+1} + \frac{3}{n+2} - \frac{1}{n+3}\right) $
$\displaystyle \frac{1}{n(n+1) (n+2)(n+3)(n+4)}= \frac{1}{4!}\left(\frac{1}{n}-\frac{4}{n+1} + \frac{6}{n+2} - \frac{4}{n+3} + \frac{1}{n+4}\right) $
Then one can group the terms as to make it a telescopic series. Just my 2 cents.
This is similar to Krizalid's post except I am not sure this is what he had in mind
Consider
$\displaystyle \frac{1}{1-x}=\sum_{n=0}^{\infty}x^n~~|x|<1$
I dont feel like justifying interchanging the integral and summation, but just notice that your sum
$\displaystyle \begin{aligned}\sum_{n=1}^{\infty}\frac{2n-1}{n(n+1)(n+2)}&=\sum_{n=1}^{\infty}\frac{2}{(n+1) (n+2)}-\sum_{n=1}^{\infty}\frac{1}{n(n+1)(n+2)}\\
&=2\int_0^1\left\{\int\frac{dx}{1-x}\right\}-\int_0^1\left\{\iint\frac{dx}{x(1-x)}\right\}\\
&=\frac{3}{4}\end{aligned}$
It is obvious that any integral or derivative of a power series evaluated at its center (in this case 0) is 0. So instead let us consider
$\displaystyle \begin{aligned}\iiint\frac{1}{x(1-x)}\bigg|_{x=1}&=\iiint\sum_{n=1}^{\infty}x^{n-1}\bigg|_{x=1}\\
&=\sum_{n=1}^{\infty}\iiint x^{n-1} dx|_{x=1}\\
&=\sum_{n=0}^{\infty}\frac{x^{n+2}}{n(n+1)(n+2)}\b igg|_{x=1}\\
&=\sum_{n=0}^{\infty}\frac{1}{n(n+1)(n+2)}\end{ali gned}$
Is that what you mean?