Let $\text{Si} (x) = \int_0^x \frac{\sin t}{t} d t$.
- Determine the Taylor series for $\text{Si}$ about $0$ and prove that it converges to $\text{Si}$.
- Show that $\text{Si} (\pi) > \frac{\pi}{2}$. You may use that $\pi > 3$ and $\pi^2 < 10$.
I used the Taylor series of $\sin$ to derive $\text{Si} (x) = \sum_{n = 0}^\infty \frac{(-1)^n x^{2 n + 1}}{(2 n + 1) (2 n + 1)!}$. Here are my attempts to answer the questions and the issues I'm struggling with:
- How can I show that the series converges to $\text{Si}$? Do I need the $n$th derivative of $\text{Si}$? I showed that, with $a_n := \frac{(-1)^n x^{2 n + 1}}{(2 n + 1) (2 n + 1)!}$, there is for any $x \in \mathbb{R}$ an $N \in \mathbb{N}$, so that $|a_{n + 1}| < |a_n|$ for $n \ge N$: $$ \left| \frac{a_{n + 1}}{a_n} \right| = \frac{(2 n + 1) |x|^2}{(2 n + 3)^2 (2 n + 2)} < \frac{(2 n + 2) |x|^2}{(2 n + 3)^2 (2 n + 2)} = \frac{|x|^2}{(2 n + 3)^2}. $$ I conclude with the ratio and Leibniz test that the series converges. This doesn't seem to imply, though, that the series converges to $\text{Si}$. I doubt therefore that my prove addresses the question.
- I guess that I'm supposed to use the Taylor series to show that $\text{Si} (\pi) > \frac{\pi}{2}$. With my notation, $a_{n + 1} < a_n$ for $n > 0$ for $x = \pi$, because $$ (2 \cdot 1 + 2) \pi^2 = 4 \pi^2 < 4 \cdot 10 < 5^2 \cdot 4 = (2 \cdot 1 + 3)^2 (2 \cdot 1 + 2) $$ I can then use $\pi > 3$ and $\pi^2 < 10$ to argue that $\text{Si} (\pi) = \pi - \frac{\pi^3}{3 \cdot 3!} + \frac{\pi^5}{5 \cdot 5!} - \frac{\pi^7}{7 \cdot 7!} + \sum_{n = 4}^\infty \frac{(-1)^n x^{2 n + 1}}{(2 n + 1) (2 n + 1)!} > \pi - \frac{\pi^3}{3 \cdot 3!} + \frac{\pi^5}{5 \cdot 5!} - \frac{\pi^7}{7 \cdot 7!} > \frac{\pi}{2}$. However, this involves numerical calculations that seem overly combersome.
I assume that I'm missing some points. I would appreciate any help.