# Minimum degree of $T_n(x)$ in knowing that $\mid f(\pi) - T_n(\pi) \mid \leq \frac{1}{100}$

I got the taylor expansion of $f(x) = \ \int_{\frac{\pi}{2}}^{x} \frac{\cos(t)}{t - \frac{\pi}{2}}dt$ around $a=\frac{\pi}{2}$ which is $$T(x) = \sum_{n=0}^\infty(-1)^{n+1} \frac{1}{(2n+1)!(2n+1)}(x-\frac{\pi}{2})^{2n+1}.$$

Let $T_n(x)$ the Taylor polynomial of degree $n$ of $f$ around $a=\frac{\pi}{2}$. What is the minimum $n$ of $T_n(x)$ so that $$\mid f(\pi) - T_n(\pi) \mid \leq \frac{1}{100}?$$ In fact, the answer is $n=1$ and $T_1(x) = -(x-\frac{\pi}{2})$, but I can't explain why rigorously.

Please let me know if the question is unclear.

You have that $$f(\pi) = \sum_{n=0}^\infty (-1)^{n+1} a_n, \qquad a_n := \frac{1}{(2n+1)! (2n+1)}\, \left(\frac{\pi}{2}\right)^{2n+1}\,.$$ You have that the series is alternating, and $(a_n)$ is a sequence of positive numbers converging motononically to $0$.
If we denote by $s_n := \sum_{k=0}^n (-1)^{n+1} a_n$ the $n$-th partial sum of the series, you have the following error estimate: $$|f(\pi) - s_n| \leq a_{n+1}$$ (see the proof of the Leibniz criterion of convergence).
You can now check in a moment that, for $n=1$, the error is bounded by $$a_2 = \frac{1}{7! 7} \left(\frac{\pi}{2}\right)^{7} \simeq 0.67 \cdot 10^{-3} < \frac{1}{100}\,.$$