The given mathematical series is:

$\sum_{n=0}^{\infty} \frac{x^{4n}}{2n! + 1}$

You are asked to determine whether the series converges for all $x$, using a comparison test.

Steps to analyze convergence:

1. Identify general term: The general term of the series is: $a_n = \frac{x^{4n}}{2n! + 1}$

2. Simplify and compare: For large $n$, $2n! + 1 \approx 2n!$ because the factorial term dominates over the constant. So, for large $n$, we can approximate the general term as: $a_n \approx \frac{x^{4n}}{2n!}$

3. Comparison to a known convergent series: The series $\sum_{n=0}^{\infty} \frac{x^{4n}}{n!}$ is the Taylor series for $e^{x^4}$, which converges for all $x$. Since the denominator of our given series is slightly larger (due to $2n! + 1$), the general term of the given series is smaller than that of the exponential series for large $n$.

4. Conclusion by comparison: Since the series $\sum \frac{x^{4n}}{n!}$ converges for all $x$ and the terms of the given series are smaller for large $n$, by the comparison test, the original series also converges for all $x$.

Would you like further details on any of the steps?

Follow-up Questions:

1. What is the comparison test in series convergence?
2. Why does $2n! + 1$ simplify to $2n!$ for large $n$?
3. How does the series $e^{x^4}$ relate to the given problem?
4. What are other common convergence tests for infinite series?
5. Can we use the ratio test here, and what would it show?

Tip: In comparison tests, always aim to compare with a well-known series like geometric or exponential series, which have clear convergence properties.