Power series that satisfies $f^{(k)}(0) = (k!)^2$

Question

To solve a larger problem I must solve the sub-problem of finding a power series that satisfies: $$f^{(k)}(0) = (k!)^2.$$ My textbook states that this is "obviously" $$\sum^{\infty}_{k=0}k!z^k.$$ Why is this?

Answer 1

For any power-series $\sum_{k=0}^{\infty}a_k x^k$ one has that $f^{(k)}(0)= k! a_k$.

This is true in a 'formal' sense for any sequence of coefficients, and it is true in the analytic sense of derivative if the power-series has positive radius of convergence.

Answer 2

Hint $$f(z)=\sum_{n=0}^{\infty} a_n z^n\implies a_n= \frac{f^{(n)}(0)}{n!} $$

Answer 3

Disclaimer: This does not answer the posed question (which has already been answered) but it does explain how to convert your non-convergent power series into a function satisfying your differential equations.

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Your series doesn't converge for any $z\ne0$. Instead, consider the Borel sum of your series:

$$\mathcal B\sum_{n=0}^\infty n!z^n=\int_0^\infty\exp(-t)\left(\sum_{n=0}^\infty z^nt^n\right)~\mathrm dt=\int_0^\infty\frac{\exp(-t)}{1-zt}~\mathrm dt=\frac1z\exp\left(-\frac1z\right)\operatorname{Ei}\left(\frac1z\right)$$

when $\Re(z)<0$ and where $\operatorname{Ei}(x)$ is the principle exponential integral.

$$\operatorname{Ei}(x)=P.V.\int_{-x}^\infty-\frac{e^{-t}}t~\mathrm dt$$

This function happens to satisfy

$$\lim_{z\to0}\frac{\mathrm d^n}{\mathrm dz^n}\frac1z\exp\left(-\frac1z\right)\operatorname{Ei}\left(\frac1z\right)=(n!)^2$$

As intended.