TITLE: The $n$th derivative of $\frac{1}{5x^5 + 1}$ at $x=0$ is $0$ for all $n$ which is not a multiple of $5$?
QUESTION [5 upvotes]: How can I show this? This is part of an assigned task for which I am to find the Taylor Series expansion of $f(x) = 1 / (5x^5 + 1)$ around $0$. I have noticed by studying the $n$th derivatives of this function that for $n$ is not a multiple of $5$, the derivative evaluates to $0$.
I believe this is useful but how can I prove this? I believe it may be related to the fact that odd derivatives of even functions at $x=0$ are $0$. Is there a theorem for the function that is $1/(5x^5 + 1)$ that says its derivatives evaluate to $0$ around $0$?
Cheers!

REPLY [2 votes]: We can write
$$
f(x)=\frac{1}{1-(-\sqrt[5]5x)^5}=\sum_{n=0}^\infty (-1)^{5n} 5^nx^{5n}
$$
by the geometric series for $|x|<1/\sqrt[5]5$. In particular the coefficients of the series are $f^{(n)}(0)/n!$ and so $f^{(n)}(0)$ is always a multiple of $5$ for all n.