I'm struggling with the following inequality. In particular, I don't understand how to demonstrate that this term is larger than the left hand side for all $n \in \mathbb{N}$. It seems to be another term for the sequence but I don't have the intuition for its growth.
Main Question: Show that $$\left| \tan^{-1}(x) -\left(x - \frac{x^3}{3} + \frac{x^5}{5} - \dots + (-1)^{n-1}\frac{t^{2n-1}}{2n-1 } \right) \right| \leq \frac{x^{2n+1}}{2n+1}$$
For context, I am trying to show an approximation of $\pi$. The steps for the proof are as follows:
Show that $$\frac{1}{1+t^2} = 1 - t^2 + t^4 - \dots +(-1)^{n-1}t^{2n-2} + (-1)^n \frac{t^{2n}}{1+t^2}$$
Integrate both sides to show that $$\tan^{-1} (x) = x - \frac{x^3}{3}+\frac{x^5}{5} - \dots +\frac{(-1)^{n-1}x^{2n-1}}{2n-1} + \int_0^{x} \frac{(-1)^nt^{2n}}{1+t^2}dt$$
Show that $$\left| \tan^{-1}(x) -\left(x - \frac{x^3}{3} +\frac{x^5}{5} - \dots + (-1)^{n-1}\frac{t^{2n-1}}{2n-1 } \right) \right| \leq \frac{x^{2n+1}}{2n+1}$$
Show that $$\tan^{-1}\frac{1}{2} + \tan^{-1}\frac{1}{3} \frac{\pi}{4}$$
Finally show that $$\left|\frac{\pi}{4} - \left(\left(\frac{1}{2} +\frac{1}{3}\right) - \frac{1}{3}\left(\frac{1}{2^2} + \frac{1}{3^2}\right) + \frac{1}{5}\left(\frac{1}{2^5} + \frac{1}{3^5}\right) - \cdots + \frac{(-1)^{n-1}}{2n-1} \left(\frac{1}{2^{2n-1}} + \frac{1}{3^{2n-1}}\right)\right) \right| \leq \frac{1}{n\cdot2^{2n-1}} $$
Here are some hints.
I assume you have already shown that $$\frac{1}{1+t^2} = 1 - t^2 + t^4 - \dots +(-1)^{n-1}t^{2n-2} + (-1)^n \frac{t^{2n}}{1+t^2}.$$
This implies that $$\left|\frac{1}{1+t^2} -\left( 1 - t^2 + t^4 - \dots +(-1)^{n-1}t^{2n-2} \right)\right| = \frac{t^{2n}}{1+t^2} \le t^{2n}.$$
That is, $$-t^{2n}\le \frac{1}{1+t^2} -\left( 1 - t^2 + t^4 - \dots +(-1)^{n-1}t^{2n-2} \right)\le t^{2n}.$$
Now try integrating this inequality.