Why doesn't telescoping change the nature of the infinite series Finding the value of $\frac{1}{n(n+1)}$ as $n\rightarrow \infty$ doesn't change the value $L$ of the limit of the series, if we find the limit by telescoping the series by $\frac{1}{n} - \frac{1}{n+1}$.
Why does this happen if it follows that we cannot assign any definite meaning to the sum of an infinite series of terms, without first ordering them in some definite manner.
 A: Remember that
$$
\sum_{n=1}^\infty\frac1{n(n+1)}=\lim_{m\to\infty}\sum_{n=1}^m\frac1{n(n+1)}
$$
Now we have converted the series to the limit of a sequence of finite sums. With finite sums, we don't have to worry about conditional convergence. In most cases where there is a conditionally convergent series, or here where a pair of divergent series, involved, it is best to convert the series to a limit of a sequence of finite sums.
Next, expand the term using partial fractions and perform the telescoping:
$$
\begin{align}
\sum_{n=1}^m\frac1{n(n+1)}
&=\sum_{n=1}^m\left(\frac1n-\frac1{n+1}\right)\\
&=\sum_{n=1}^m\frac1n-\sum_{n=2}^{m+1}\frac1n\\
&=1+\sum_{n=2}^m\frac1n-\sum_{n=2}^m\frac1n-\frac1{m+1}\\
&=1-\frac1{m+1}
\end{align}
$$
All the sums just above are finite, so all the manipulations we've done are good. Now, we just need to take the limit as $m\to\infty$:
$$
\begin{align}
\sum_{n=1}^\infty\frac1{n(n+1)}
&=\lim_{m\to\infty}\left(1-\frac1{m+1}\right)\\
&=1
\end{align}
$$
A: There is a famous theorem of Riemann that says that for a conditionally convergent series, we can re-order the terms so that the series converges to any limit or even diverges.
A series $a_1 + a_2 + a_3 +\cdots$ is said to converge absolutely if the series $|a_1|+|a_2|+|a_3|+\cdots$ converges. If a covergent series does not converge absolutely then it converges conditionally. 
For an absolutely convergent series you can reorder the terms however you like and you will always get the same limit. It is only for conditionally convergent series that the order matters.
In the example you give, $n \ge 1$ and so we have
$$\frac{1}{n(n+1)} = \left|\frac{1}{n(n+1)}\right|$$
That means that the series converges if, and only if, it converges absolutely. That means that if the series has a limit, say $L$, then all limits of all possible re-ordered series will also be $L$. 
