Need help proving inequality for nth derivative of function

Question

I am currently working on some proofs for a problem sheet for my real analysis course which has to do with Taylor's Theorem. I am stuck on this one and have not had any luck so far. I am hoping that someone could maybe give me a hint in the right direction on this:

1. Show that $\forall x \in R$, $|\frac{d^n}{dx^n} \frac{1}{1+x^2}| \leq n!$

So far with this problem, I have expanded the function into partial fractions using the factors $(1+xi)(1-xi)$ to get

$\frac{1}{1+x^2} = \frac{1}{2} (\frac{1}{1+xi} + \frac{1}{1-xi})$

I have tried expanding $\frac{1}{1+xi}$ and $\frac{1}{1-xi}$ using Taylor and I was able to derive the series expansion for $\frac{1}{1+x^2}$, but have not been able to make any progress on the actual proof. I am not sure where to go or if/how I can use the Taylor expansions of $\frac{1}{1+xi}$ and $\frac{1}{1-xi}$. Am I going in the right direction? Any tips? Thanks!

Answer 1

You don't need Taylor expansion (this will give you derivatives only in $0$), you can just explicitly find derivative as function of $x$.

We have $\frac{d^n}{dx^n} \frac{1}{1 + \alpha x} = \frac{(-\alpha)^n n!}{(1 + \alpha x)^{n + 1}}$ (easily provable by induction).

Thus, $$\left|\frac{d^n}{dx^n}\frac{1}{1 + x^2}\right| \leq \\ \frac{1}{2}\left(\left|\frac{d^n}{dx^n}\frac{1}{1 + ix}\right| + \left|\frac{d^n}{dx^n}\frac{1}{1 - ix}\right|\right) = \\ \frac{1}{2}\left(\left|\frac{(-i)^n n!}{(1 + i x)^{n + 1}} \right| + \left|\frac{i^n n!}{(1 - ix)^{n + 1}} \right|\right) \leq \\ \frac{1}{2} (n! + n!) = n! $$