Hard Telescoping Series Finding the explicit sum of a telescoping series with two factors in the denominator is quite easy: we split the fractions in the difference of two subpieces.
But what about 2+ factors? E.g., consider:
$$\sum\frac{1}{(2n+1)(2n+3)(2n+5)}$$
We could split it in three pieces by partial fractions, but difference of three pieces would be useless.
Suggestions? Thanks.
 A: First, note that the telescoping series method only works on certain fractions. In particular, in order for the fractions to cancel out, we need the numerators to be the same. The typical example of telescoping series (for partial fractions) is 
$$ \frac {1}{n(n+1)} = \frac {1}{n} - \frac {1}{n+1} \Rightarrow \sum_{i=1}^n \frac {1}{i(i+1)} = \sum_{i=1}^n \frac {1}{i} - \frac {1}{i+1} = \frac {1}{1} - \frac {1}{n+1}$$
If the numerators do not cancel out completely, then telescoping series will no longer work. For example, since we have
$$ \frac {2}{n} - \frac {1}{n+1} = \frac {n+2}{n(n+1)},$$
We find that $\sum \frac {n+2}{n(n+1)}$ doesn't yield to the telescoping series method.
How do we extend this? The key idea is that the numerators must cancel out. For example, from $1-2+1=0$, we can create
$$ \frac {1}{n} - \frac {2}{n+1} + \frac {1}{n+2} = \frac {2}{n(n+1)(n+2)}$$
which then tells us that 
$$\sum_{i=1}^n \frac {2}{i(i+1)(i+2)} = \sum_{i=1}^n\frac {1}{i} - \frac {2}{i+1} + \frac {1}{i+2} = \frac {1}{1} - \frac {1}{2} + \frac {1}{n+1} - \frac {1}{n+2}.$$
You can easily see how to create more telescoping series sums using this idea. For example, what can we do with $3-4+1 = 0$ and the fraction:
$$ \frac 3 n - \frac 4 {n+1} + \frac 1 {n+2} = \frac {2(n+3)}{n(n+1)(n+2)}?$$

Edit: So, it sounds to me that you do not really see how telescoping series works. Let me expand it out, so you could understand it better. Let's first take the typical example of $\sum \frac {1}{i(i+1)}$. We can see that $\frac {1}{i(i+1)} = \frac {1}{i} - \frac {1}{i+1}$. What does this tell us? We have
$$\begin{array}{llllllllll}
\frac {1}{1 \times 2} & = & \frac {1}{1} & - \frac {1}{2} \\
\frac {1}{2 \times 3} & = & & +\frac {1}{2} & - \frac {1}{3} \\
\frac {1} {3 \times 4} & = & & & + \frac {1}{3} & - \frac {1}{4} \\
\vdots & = \vdots \\
\frac {1}{(n-1) \times n} = &&&&&&& +\frac {1}{n-1} &- \frac {1}{n}\\
\frac {1}{n \times (n+1)} = &&&&&&& +\frac {1}{n} &- \frac {1}{n+1}\\
\end{array}$$
Now, summing up the LHS, we get $\sum_{i=1}^n \frac {1}{i(i+1)}$ as intended. Let's sum up the RHS according to each column. Then, we clearly see that lots of things cancel out, leaving us with $\frac 1 1 - \frac 1 {n+1}$.
We know do the same with $\frac {1}{n} - \frac {2}{n+1} + \frac {1}{n+2} = \frac {2}{n(n+1)(n+2)}$. Let's write it out as:
$$\begin{array}{llllllllll}
\frac {2}{1 \times 2 \times 3} & = & \frac {1}{1} & - \frac {2}{2} & + \frac {1}{3}\\
\frac {2}{2 \times 3 \times 4} & = & & +\frac {1}{2} & - \frac {2}{3} & + \frac {2}{4}\\
\frac {2} {3 \times 4 \times 5} & = & & & + \frac {1}{3} & - \frac {2}{4} & + \frac {1}{5} \\
\vdots & = \vdots \\
\frac {2}{n \times (n+1) \times (n+2)} = &&&&&&& +\frac {1}{n-1} &- \frac {2}{n} &+ \frac {1}{n+1}\\
\frac {2}{n \times (n+1) \times (n+2)} = &&&&&&&& +\frac {1}{n} &- \frac {2}{n+1} + \frac {1}{n+2}\\
\end{array}$$
Now, can you tell what we get when we sum up the RHS over the columns? You should see that we get $\frac {1}{1} - \frac {1}{2} - \frac {1}{n+1} + \frac {1}{n+2}$.
A: $$\dfrac1{(2n+1)(2n+3)(2n+5)} = \dfrac18 \left( \dfrac1{2n+1} - \dfrac1{2n+3} \right) -\dfrac{1}{8}\left( \dfrac1{2n+3} - \dfrac1{2n+5} \right)$$
Each bracket is telescopic.
The reason why it (always) turns into a telescopic sum, is simple to understand: check the answer by Pambos. It is actually the same thing what you get if you decompose each of his fractions into partial fractions. But as soon as you realize that this is the case, all you need is to do the complete PFD and group them the right way...
A: Write $$\dfrac1{(2n+1)(2n+3)(2n+5)} =\\ \dfrac1{2n+3} \left(\frac{1}{(2n+1)(2n+5)}\right)
=\\ \dfrac1{4(2n+3)} \left(\frac{1}{2n+1}-\frac{1}{2n+5}\right)
=\\ \dfrac14 \left(\frac{1}{(2n+1)(2n+3)}-\frac{1}{(2n+3)(2n+5)}\right)
=a_n-a_{n+1},\\
\text{for  } \ a_n=\dfrac14 \left(\frac{1}{(2n+1)(2n+3)}\right)
$$
to make it easy (telescopic with two factors).
A: A much better method than other answers:
Write an= 1/(2n+1)(2n+3)(2n+5)=  1/4(2n+1)(2n+3) - 1/4(2n+3)(2n+5)
Now if you consider f(n)= 1/4(2n+1)(2n+3), then an= f(n)-f(n+1).
Which telescopes.
