There are many series to calculate in Calculus. In particular, Taylor Series are widely used in Calculus courses. In this post, I’ll explain how to use Taylor Series to calculate the following alternating series:
$$ 1 \, – \, \frac{1}{2} + \frac{1}{3} \, – \, \frac{1}{4} + \frac{1}{5} \, – \, \frac{1}{6} + \cdots $$
Notice that
$$ 1 \; – \, \frac{1}{2} + \frac{1}{3} \, – \, \frac{1}{4} + \frac{1}{5} \, – \, \frac{1}{6} + \cdots $$
$$= 1 \; – \, \frac{1^2}{2} + \frac{1^3}{3} \, – \, \frac{1^4}{4} + \frac{1^5}{5} \, – \, \frac{1^6}{6} + \cdots $$
$$= (2 \; – \, 1) \; – \, \frac{(2 \; – \, 1)^2}{2} + \frac{(2 \; – \, 1)^3}{3} \, – \, \cdots $$
It turns out that there is a nice Taylor series for the natural logarithm:
$$ \ln(x) = (x \; – \, 1) \; – \, \frac{(x \; – \, 1)^2}{2} + \frac{(x \; – \, 1)^3}{3} \, – \, \cdots $$
This is known as the Taylor series for the natural logarithm at \( 1 \). If we substitute \( 2 \) for \( x \) in the above Taylor series, we get:
$$ \ln(2) = (2 \; – \, 1) \; – \, \frac{(2 \; – \, 1)^2}{2} + \frac{(2 \; – \, 1)^3}{3} \, – \, \cdots $$
So,
$$ 1 \, – \, \frac{1}{2} + \frac{1}{3} \, – \, \frac{1}{4} + \frac{1}{5} \, – \, \frac{1}{6} + \cdots = \ln(2) $$
Using a calculator, we find that \( \ln(2) \) is around
$$ 0.693 $$
So we can say that the above alternating series is around that value.
