https://file.aiquickdraw.com/mathbot/file/17305594802573t3na1c3.jpg

The series given in the image is:

\[
\sum_{n=0}^{\infty} \frac{(-1)^n x^{2n}}{(2n)!}
\]

To determine if this series is absolutely convergent, we can examine the absolute value of each term:

\[
\left| \frac{(-1)^n x^{2n}}{(2n)!} \right| = \frac{|x|^{2n}}{(2n)!}
\]

This series resembles the Taylor series for \( e^{x^2} \), which is known to converge for all real values of \( x \). Since the factorial in the denominator grows much faster than the powers of \( x \) in the numerator, the series converges absolutely for all \( x \). 

Would you like further details on the convergence test used here, or have any other questions?

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1. How can we use the Ratio Test to confirm absolute convergence here?
2. What is the general form of a Taylor series for exponential functions?
3. Why does the factorial growth in the denominator ensure convergence?
4. What does it mean for a series to converge "absolutely" versus "conditionally"?
5. Can we approximate the sum of this series for small values of \( x \)?

**Tip:** In series convergence problems, factorials in the denominator generally indicate rapid convergence due to the factorial’s growth rate.

Determine whether the series \( \sum_{n=0}^{\infty} \frac{(-1)^n x^{2n}}{(2n)!} \) is absolutely convergent or divergent.

Explore the convergence of the series \( \sum_{n=0}^{\infty} \frac{(-1)^n x^{2n}}{(2n)!} \), using absolute convergence criteria and factorial growth analysis. This calculus-level problem demonstrates why the series converges absolutely for all real values of \( x \).

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