# Radius of convergence of, $f(z)=10+\sum_{n=1}^\infty \frac{p(n)}{n!}z^{n!}$

## Problem statement

If $$p$$ is a polynomial with positive integer coefficients, what is the radius of convergence of, $$\displaystyle f(z)=10+\sum_{n=1}^\infty \frac{p(n)}{n!}z^{n!}$$

### My Attempt

The value of the constant 10 is irrelevant. I considered the term $$\frac{p(n)}{n!}$$.

Since any arbitrary order positive polynomial will always increase slower than $$n!$$, I made the assumption that the denominator increases asymptotically faster than the numerator and thus always converges for all values. And the radius of convergence must be infinity when applying the root test.

This was a problem from my complex analysis assignment, and the correct answer is apparently a radius of 1 not infinity.

How do I go about solving this problem correctly?

Thanks for any and all help!

• Well, for example $2^{n!}$ grows a lot faster than $n!$ Aug 30, 2021 at 3:28
• A common mistake in dealing with this type of power series is to treat $p(n)/n!$ as the coefficient of $z^n$, instead of $z^{n!}$. Aug 30, 2021 at 4:36

We consider first the case where $$p(z)=z^m$$, then $$p(n?)^{1/n}=(n?)^{m/n}\leq (n^{1/n})^m$$. Moreover $$1=1^{m/n}\leq n?^{m/n}=p(n?)^{1/n}$$. Where $$n?$$ denotes the inverse factorial (or inverse of $$\Gamma(n+1)$$). What we have shown is that the sequence $$p(n?)^{1/n}$$ tends to $$1$$ as $$n$$ tends to $$\infty$$.

We then get $$\lim_{n\to\infty} \frac{p(n?)^{1/n}}{n^{1/n}}=\frac{1}{1}=1.$$

It now also follows that if $$a_n$$ is the sequence $$p(n?)/n$$ if $$n?$$ is an integer and $$0$$ otherwise, then

$$\limsup_{n\to\infty} a_n^{1/n}=1$$

which is exactly what we need. Hence the radius of convergence of $$\sum_{n=1}^{\infty} a_nz^n=\sum_{n=0}^{\infty} \frac{p(n)}{n!}z^{n!}$$ is $$1$$. Now all we have to do to finish the proof for any polynomial, is to realize that for large $$|z|$$ we have $$|p(z)|=|a_mz^m+a_{m-1}z^{m-1}+...+a_1z+a_0|\leq |a_mz^m|+|a_{m-1}z^{m-1}|+...+|a_1z|+|a_0|\leq$$$$|a_mz^m|+|a_{m-1}z^m|+...+|a_1z^m|+|a_0z^m|=(|a_m|+...+|a_0|)|z|^m.$$

The estimate now follows similarly with $$C^{1/n}\to 1$$ for any nonzero constant.