text: # Need help proving inequality for nth derivative of function
  
  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!
  
  • Is there any condition on n? Commented Apr 3, 2023 at 19:46
  
  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, $$\left|\frac{d^n}{dx^n}\frac{1}{1 + x^2}\right| \leq \\ \frac{1}{2}\left(\left|\frac{d^n}{dx^n}\frac{1}{1 + ix}\right| + \left|\frac{d^n}{dx^n}\frac{1}{1 - ix}\right|\right) = \\ \frac{1}{2}\left(\left|\frac{(-i)^n n!}{(1 + i x)^{n + 1}} \right| + \left|\frac{i^n n!}{(1 - ix)^{n + 1}} \right|\right) \leq \\ \frac{1}{2} (n! + n!) = n!$$
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