TITLE: Alternating Series and Convergence
QUESTION [6 upvotes]: The question is: Approximate the value of the series within an error of at most $10^{-4}$.
$$ \sum_{n=1}^\infty \frac{(-1)^{(n+1)}}{(n+79)(n+73)} $$
According to 
$$|S_N - S| ≤ a_{N+1}$$
what is the smallest value of $N$ that approximates $S$ to within an error of at most $10^{-4}$?
I have tried the following to get $N$:
$$a_n = \frac{1}{(n+79)(n+73)}$$
$$\frac{1}{(n+79)(n+73)} < \frac{1}{10000}$$
$N=25$
How do I solve for $S$?

REPLY [3 votes]: As suggested by vonbrand, the series can be split by using partial fraction decomposition. Rewrite
$$
\begin{align}
\sum_{n=1}^\infty \frac{(-1)^{(n+1)}}{(n+79)(n+73)}&=\sum_{n=1}^\infty \frac{(-1)^{(n+1)}}{6}\left(\frac{1}{n+73}-\frac{1}{n+79}\right).
\end{align}
$$
This is a telescoping series and the fraction terms will cancel out for $n>6$. Thus
$$
\begin{align}
\sum_{n=1}^\infty \frac{(-1)^{(n+1)}}{(n+79)(n+73)}&=\frac{1}{6}\sum_{n=1}^\infty \left(\frac{(-1)^{(n+1)}}{n+73}-\frac{(-1)^{(n+1)}}{n+79}\right)\\
&=\frac{1}{6}\sum_{n=1}^6 \frac{(-1)^{(n+1)}}{n+73}.\\
\end{align}
$$
Since the series is only left $6$ terms, its summation value will easy be obtained by pocket calculator.

$$\large\color{blue}{\text{# }\mathbb{Q.E.D.}\text{ #}}$$