Mclaurin on $\arccos(\frac{n^2-1}{n^2+1})$

Question

I have expanded $\lim_{n\to \infty} \arccos(\frac{n^2-1}{n^2+1})$ to $\arccos(1-\frac{2}{n^2})$ and now i dont know what to do. I wrote the function on walfram alpha and he told me that the result is $\frac{2}{n}+\frac{1}{3n^3}+O((\frac{1}{n})^6)$ you can explain to me how that result came out from where?

Answer 1

Actually, for $x\in [0,+\infty[$,

$$\arccos \frac{1-x^2}{1+x^2}=2\arctan x$$

To see why, write $x=\tan \theta$ for $\theta \in[0,\pi/2[$, then $\theta=\arctan x$ and

$$\frac{1-x^2}{1+x^2}=\frac{1-\tan^2 \theta}{1+\tan^2 \theta}=\frac{\cos^2\theta -\sin^2\theta}{\cos^2\theta +\sin^2\theta}=\cos 2\theta$$

And since $2\theta \in [0,\pi[$, $\arccos(\cos2\theta)=2\theta$

Now, for $|x|<1$

$$\arctan x=\sum_{n=0}^\infty (-1)^n \frac{x^{2n+1}}{2n+1}$$

And (don't forget to rename the sum variable)

$$\arccos \frac{n^2-1}{n^2+1}=\arccos \frac{1-1/n^2}{1+1/n^2}=2\sum_{k=0}^\infty \frac{(-1)^k}{(2k+1)n^{2k+1}}$$

By the way, since the LHS in the first equation is even, and the RHS is odd, you get, for any $x \in \Bbb R$,

$$\arccos \frac{1-x^2}{1+x^2}=2|\arctan x|$$

One way to "guess" the result in the first place, is to remember the half-angle formula: if $x=\tan (\theta/2)$, then $\cos \theta = \frac{1-x^2}{1+x^2}$.

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Edit, for malloc

Here is a slightly different proof.

In the above, I apply the development of $\arctan$ on $]-1,1[$, and I let $x=1/n$.

But, still using

$$\arccos \frac{1-x^2}{1+x^2}=2\arctan x$$

And with $x=n$, we can also use the development of $\arctan$ at $\infty$. For this, you will need, for $x>0$:

$$\arctan x + \arctan \frac 1 x = \frac {\pi}2$$

Thus if $n=x >1$ (so that the development of $\arctan \frac 1x$ holds),

$$\arctan n = \frac {\pi}2-\arctan \frac 1 n = \frac{\pi}{2}-\sum_{k=0}^\infty (-1)^k \frac{(\frac{1}{n})^{2k+1}}{2k+1}$$

And finally (since $\arccos (-x)=\pi-\arccos(x)$),

$$\arccos \frac{n^2-1}{n^2+1}=\pi-\arccos \frac{1-n^2}{1+n^2}$$ $$=\pi-2\arctan n=\pi-2\left(\frac{\pi}{2}-\sum_{k=0}^\infty (-1)^k \frac{(\frac{1}{n})^{2k+1}}{2k+1}\right)$$

$$\arccos \frac{n^2-1}{n^2+1}=2\sum_{k=0}^\infty (-1)^k \frac{(\frac{1}{n})^{2k+1}}{2k+1}$$

Answer 2

$\lim_{n\to \infty}= arccos{\frac{n^2-1}{n^2+1}}=arccos\left({1-\frac{2}{n^2+1}}\right)$

So as $n\to\infty$ this goes to $arccos(1)$ which is equal to $0$.