1. **State the problem:** We want to understand and analyze the series $$\sum_{n=-0.2}^{10} \frac{(2n)!}{4^n (n!)^2 (2n+1)} x^{2n+1}$$. 2. **Important note:** The summation index $n$ must be an integer, but here it starts at $-0.2$, which is not an integer. Typically, summations over factorial expressions require integer indices starting from 0 or a positive integer. 3. **Assuming the intended sum is from $n=0$ to $10$:** The series resembles the Taylor series expansion of the inverse sine function: $$\arcsin(x) = \sum_{n=0}^\infty \frac{(2n)!}{4^n (n!)^2 (2n+1)} x^{2n+1}$$ 4. **Explanation:** - The factorial terms and powers match the coefficients of the arcsin series. - The series converges for $|x| \leq 1$. 5. **Therefore, the partial sum from $n=0$ to $10$ is an approximation of $\arcsin(x)$:** $$\sum_{n=0}^{10} \frac{(2n)!}{4^n (n!)^2 (2n+1)} x^{2n+1} \approx \arcsin(x)$$ 6. **Summary:** The given series (corrected to start at $n=0$) is the Taylor polynomial of degree 21 for $\arcsin(x)$. Final answer: $$\sum_{n=0}^{10} \frac{(2n)!}{4^n (n!)^2 (2n+1)} x^{2n+1} = \arcsin(x) \text{ (approximation)}$$