Question

For each power series use the result of Exercise 4 to find the radius of convergence \(R\). If \(R>0\), find the open interval of convergence. (a) \(\sum_{m=0}^{\infty} \frac{(-1)^{m}}{(27)^{m}}(x-3)^{3 m+2}\) (b) \(\sum_{m=0}^{\infty} \frac{x^{7 m+6}}{m}\) (c) \(\sum_{m=0}^{\infty} \frac{9^{m}(m+1)}{(m+2)}(x-3)^{4 m+2}\) (d) \(\sum_{m=0}^{\infty}(-1)^{m} \frac{2^{m}}{m !} x^{4 m+3}\) (e) \(\sum_{m=0}^{\infty} \frac{m !}{(26)^{m}}(x+1)^{4 m+3}\) (f) \(\sum_{m=0}^{\infty} \frac{(-1)^{m}}{8^{m} m(m+1)}(x-1)^{3 m+1}\)

Short answer

Answer: The interval of convergence for the given power series is \([0, 6]\).

Step-by-step solution

Apply Ratio Test

Use the Ratio Test to find the radius of convergence: \(\lim_{m \to \infty} \left|\frac{a_{m+1}}{a_m}\right| = L\) First, find the general term \(a_m = \frac{(-1)^{m}}{(27)^{m}}(x-3)^{3 m+2}\) Calculate the general term \(a_{m+1}\) by replacing \(m\) with \(m+1\): \(a_{m+1} = \frac{(-1)^{m+1}}{(27)^{m+1}}(x-3)^{3(m+1)+2}\) Now, find the ratio \(\frac{a_{m+1}}{a_m}\): \(L = \left|\frac{(-1)^{m+1}(x-3)^{3m+5}}{(27)^{m+1}} \cdot \frac{(27)^{m}}{(-1)^{m}(x-3)^{3m+2}}\right| = \frac{|x-3|^3}{27}\) The power series converges if \(L < 1\), so \(\frac{|x-3|^3}{27} < 1\).

Find the radius of convergence (R)

Solve the inequality for R: \(\frac{|x-3|^3}{27} < 1 \Rightarrow |x-3|^3 < 27\). Taking the cube root of both sides: \(|x-3| < 3\). Thus, the radius of convergence, R, is 3.

Find the open interval of convergence

The interval of convergence is given by \((3-R, 3+R) = (3-3, 3+3) = (0, 6)\). Now, check the convergence at the endpoints \(x=0\) and \(x=6\). At \(x=0\), the power series becomes \(\sum_{m=0}^{\infty} \frac{(-1)^{m}}{(27)^{m}}(-3)^{3 m+2}\). This power series converges by the Alternating Series Test. Replace \(x\) with \(6\): \(\sum_{m=0}^{\infty}\frac{(-1)^m}{(27)^m}(3)^{3m+2}\). This series also converges by the Alternating Series Test. Since both endpoints converge, the power series converges on the closed interval \((0, 6)\).