Activity 8.6.2

In this activity we approximate the values of several different alternating series using the Alternating Series Estimation Theorem.

1. Use the fact that $\sin(x) = x - \frac{1}{3!}x^3 + \frac{1}{5!}x^5 - \cdots$ to estimate $\sin(1)$ to within $0.0001$. Do so without entering “$\sin(1)$” on a computational device. After you find your estimate, enter “$\sin(1)$” on a computational device and compare the results.

2. Recall our recent work with $\int_0^1 e^{-x^2} \, dx$ which can be expressed as the series $$\begin{gathered} \int_0^1 e^{-x^2} \, dx = 1 - \frac{1}{3} + \frac{1}{2! \cdot 5} - \frac{1}{3! \cdot 7} + \cdots + \frac{(-1)^n}{n! \cdot (2n+1)} + \cdots \end{gathered}$$. Use this series representation to estimate $\int_0^1 e^{-x^2} \, dx$ to within $0.0001$. Then, compare what a computational device reports when you use it to estimate the definite integral.

3. Find the Taylor series for $\cos(x^2)$ and then use the Taylor series and to estimate the value of $\int_0^1 \cos(x^2) \, dx$ to within $0.0001$. Compare your result to what a computational device reports when you use it to estimate the definite integral.

4. Recall we know that if $|x| \lt 1$, then $$\begin{gathered} \ln(1+x) = x - \frac{1}{2}x^2 + \frac{1}{3}x^3 - \cdots + (-1)^{n-1} \frac{1}{n} x^n + \cdots \end{gathered}$$. What happens if $x = 1$?

   Explain why the series $1 - \frac{1}{2} \cdot 1^2 + \frac{1}{3} \cdot 1^3 - \cdots + (-1)^{n-1} \frac{1}{n} \cdot 1^n + \cdots$ must converge and estimate its sum to within $0.01$. What is the exact sum of this series?