New Taylor Series:
Use $\cos x = \sum_{i=0}^{\infty} (-1)^n \frac{x^{2n}}{(2n)!}$ to solve:
- Find the Taylor series of $\frac{1-\cos (2x)}{3x^2}$ at $b = 0$.
- Using the Taylor series above, compute the exact value for $\lim_{x \rightarrow 0} \frac{1-\cos (2x)}{3x^2}$. Explain your reasoning!
Exact Values:
Work backwards and recall that $\frac{1}{1+x} = \sum_{i = 0}^{\infty} (-1)^i x^i$ to find the exact value of
$$\frac{4}{5} - \frac{16}{25}+ \frac{64}{125} - \frac{256}{625}+ \cdots $$
Work backwards and recall another well known Taylor series to find the exact value of
$$1 + \frac{1}{2} + \frac{1}{4 \cdot 2!} + \frac{1}{8 \cdot 3!} + \frac{1}{16 \cdot 4!}+ \frac{1}{32 \cdot 5!} + \cdots$$