TITLE: Why doesn't telescoping change the nature of the infinite series
QUESTION [6 upvotes]: 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.

REPLY [6 votes]: 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 $m\to\infty$:
$$
\begin{align}
\sum_{n=1}^\infty\frac1{n(n+1)}
&=\lim_{m\to\infty}\left(1-\frac1{m+1}\right)\\
&=1
\end{align}
$$