<s>[INST] Help With Series (Apostol, Calculus, Volume I, Section 10.9 #9) I am looking for help finding the sum of a particular series from Apostol's Calculus (Volume I, Section 10.9, Problem 9). The trouble is that I can find the correct answer, but only using methods that aren't available at this point in the text, or with too much trouble. I would like some guidance on reducing the following series to a linear combination of convergent geometric or telescoping series, the methods by means of which I'm supposed to solve the problem. The series in question is $$\sum_{n = 1}^{\infty} \frac{(-1)^{n - 1}(2n + 1)}{n(n + 1)}$$ Decomposing the series using partial fractions and observing the series term-by-term, we see that
  $$\begin{align*}
    \sum_{n = 1}^{\infty} \frac{(-1)^{n - 1}(2n + 1)}{n(n + 1)} & ~ = ~ \sum_{n = 1}^{\infty} (-1)^{n - 1}\left(\frac{1}{n} + \frac{1}{n + 1}\right) \\
      & ~ = ~ \sum_{n = 1}^{\infty}\frac{1}{n} - \frac{1}{n + 1} \\
      & ~ = ~ 1 - \lim_{n \rightarrow \infty} \frac{1}{n} \\
      & ~ = ~ 1
  \end{align*}$$
The only problem is justifying the second line. The intuitive justification for this step is as follows
$$\begin{align*}
  & \left(\frac{1}{1} + \frac{1}{2}\right) - \left(\frac{1}{2} + \frac{1}{3}\right) + \left(\frac{1}{3} + \frac{1}{4}\right) - \left(\frac{1}{4} + \frac{1}{5}\right) + \left(\frac{1}{5} + \frac{1}{6}\right) - \left(\frac{1}{6} + \frac{1}{7}\right) + \cdots \\
  ~ = ~ & \left(\frac{1}{1} - \frac{1}{2}\right) + \left(\frac{1}{2} - \frac{1}{3}\right) + \left(\frac{1}{3} - \frac{1}{4}\right) + \left(\frac{1}{4} - \frac{1}{5}\right) + \left(\frac{1}{5} - \frac{1}{6}\right) + \left(\frac{1}{6} - \frac{1}{7}\right) + \cdots
\end{align*}$$
The problem is that I don't have any theorems about rearrangement of series at my disposal. Thus, justifying the second line devolves into demonstrating equivalence of the corresponding sequences. I'm using the test for equivalence of Cauchy sequences, while hoping it applies in this case
$$ \forall \varepsilon > 0 \exists N > 0 \forall n \geq N ~.~ \left| (-1)^{n - 1}\left(\frac{1}{n} + \frac{1}{n + 1}\right) - \left(\frac{1}{n} - \frac{1}{n + 1}\right)\right| \leq \varepsilon $$
Now this isn't particularly difficult, just tedious. However, given that no prior examples involved proving such equivalences, I assume that I'm missing something obvious that would let me transform the given series into a telescoping series. Either that, or I've messed up somewhere and need some correction. Any help would be appreciated.
 [/INST] Take a look at the sum of 
the first $m$ terms.
$\begin{align}
\sum_{n = 1}^{m} \frac{(-1)^{n - 1}(2n + 1)}{n(n + 1)}
&=\sum_{n = 1}^{m} (-1)^{n - 1}\left(\frac{1}{n} + \frac{1}{n + 1}\right)\\
&=\sum_{n = 1}^{m} (-1)^{n - 1}\frac{1}{n} + \sum_{n = 1}^{m} (-1)^{n - 1}\frac{1}{n + 1}\\
&=\sum_{n = 1}^{m} (-1)^{n - 1}\frac{1}{n} + \sum_{n = 2}^{m+1} (-1)^{n}\frac{1}{n}\\
&=1+\sum_{n = 2}^{m} (-1)^{n - 1}\frac{1}{n} + (-1)^{m+1}\frac{1}{m+1}+\sum_{n = 2}^{m} (-1)^{n}\frac{1}{n}\\
&=1-\sum_{n = 2}^{m} (-1)^{n}\frac{1}{n} + (-1)^{m+1}\frac{1}{m+1}+\sum_{n = 2}^{m} (-1)^{n}\frac{1}{n}\\
&=1+ (-1)^{m+1}\frac{1}{m+1}\\
\end{align}
$
This clearly converges to $1$.
</s>