Bound for error term in Taylor expansion of $\arctan x$.

We know

$$\begin{align*} \arctan (x) & = \int_0^x \dfrac{1}{1+t^2} dt \\ & = \int_0^x \left[1 - t^2 + t^4 - \cdots + (-1)^{n-1} t^{2(n-1)} + \dfrac{(-1)^{n} t^{2n}}{1+t^2} \right] dt\\ & = \sum_{k = 0}^{n-1} \dfrac{(-1)^k x^{2k+1}}{2k+1} + (-1)^{n}\int_0^x \dfrac{t^{2n}}{1+t^2} dt\\ & = \sum_{k = 0}^{n-1} \dfrac{(-1)^k x^{2k+1}}{2k+1} + E_{2n}(x) \end{align*}$$

Hence, we have $E_{2n} (x) = \displaystyle{(-1)^{n}\int_0^x \dfrac{t^{2n}}{1+t^2} dt}$ so we are just trying to bound the integral. So, we have

$$\begin{align*} |E_{2n} (x)| & = \left|(-1)^{n}\int_0^x \dfrac{t^{2n}}{1+t^2} dt\right|\\ & = \int_0^x \dfrac{t^{2n}}{1+t^2} dt \qquad (\text{the integrand } \geq 0 \text{ for } t \in [0,1])\\ & \leq \int_0^x t^{2n} dt \qquad \qquad \left(t \in [0,1] \implies t^{2n} \geq \dfrac{t^{2n}}{1+t^2}\right)\\ & = \dfrac{x^{2n+1}}{2n+1} \end{align*}$$

Which is the bound we were looking for.

The Lagrange formula for the remainder tends to be not useful for getting good estimates of the error.

The Taylor series for $\arctan x$ is most easily obtained by term by term integration of the series for $\frac{1}{1+x^2}$.

Note that if $-1<x<0$ or $0<x<1$, we get an alternating series. The error when you truncate at a certain point is less (in absolute value) than the (absolute value of) the first "neglected" term. That seems to be exactly what you are being asked to show.

The result about alternating series is standard, and found in most calculus books. If you need a proof, one can easily be supplied.

The same is true at $x=\pm 1$, but because of the missing power of $x$, these values should get special treatment.

Although non-rigorous, the following trick is useful.

Write

$$f'(x)=\dfrac{1}{1+x^2}=\frac{1}{2i}\left(\frac{1}{x-i}-\frac{1}{x+i}\right)$$

We have that $0<f'\leq1$

Then

$$f'(x)=-\frac{1}{2i}\left(\frac{1}{(x-i)^2}-\frac{1}{(x+i)^2}\right)$$

and successively

$$f''(x)=2!\frac{1}{2i}\left(\frac{1}{(x-i)^3}-\frac{1}{(x+i)^3}\right)$$

$$f'''(x)=-3!\frac{1}{2i}\left(\frac{1}{(x-i)^4}-\frac{1}{(x+i)^4}\right)$$

$$\cdots=\cdots$$

$$f^{(n)}(x)=(-1)^n n!\frac{1}{2i}\left(\frac{1}{(x-i)^{n+1}}-\frac{1}{(x+i)^{n+1}}\right)$$

Moreover $$\frac{1}{2i}\left(\frac{1}{(x-i)^{n+1}}-\frac{1}{(x+i)^{n+1}}\right)$$ will give a rational function in $x$ with monic polynomials where the degree in the denominator is $2n$ and in the denominator $n-1$, thus you can bound your derivatives by $0$ and $n!$