We can bound the error with,
$$
\left|R_{N}(x)\right| \leq \frac{\pi}{2}-\frac{4}{\pi} \sum_{\substack{n=1 \ \text { odd } n}}^{N} \frac{1}{n^{2}}
$$
$N=7$ is the smallest number for which our error bound is less than $10^{-1} \cdot N \geq 7$ is sufficient to make the error less that $0.1$.
$$
\left|R_{7}(x)\right| \leq \frac{\pi}{2}-\frac{4}{\pi}\left(1+\frac{1}{9}+\frac{1}{25}+\frac{1}{49}\right) \approx 0.079
$$
$N \geq 7$ is also necessary because.
$$
\left|R_{N}(0)\right|=\frac{4}{\pi} \sum_{\substack{n=N+1 \ \text { odd } n}}^{\infty} \frac{1}{n^{2}} .
$$
MATH0056 COURSE NOTES ：
It’s always OK to integrate a Fourier series term by term. We integrate the series in part (a).
$$
\begin{aligned}
&\int_{a}^{x} 1 \mathrm{~d} x \sim \frac{4}{\pi} \sum_{\substack{n=1 \
\text { odd } n}}^{\infty} \int_{a}^{x} \frac{\sin (n \xi)}{n} \mathrm{~d} x \
&x-a \sim \frac{4}{\pi} \sum_{\substack{n=1 \
\text { odd } n}}^{\infty} \frac{\cos (n a)-\cos (n x)}{n^{2}}
\end{aligned}
$$
Since the series converges uniformly, we can replace the $\sim$ with $=$.
$$
x-a=\frac{4}{\pi} \sum_{\substack{n=1 \ \text { odd } n}}^{\infty} \frac{\cos (n a)}{n^{2}}-\frac{4}{\pi} \sum_{\substack{n=1 \ \text { odd } n}}^{\infty} \frac{\cos (n x)}{n^{2}}
$$