Find all values of x for which the series inf E n=1 (3^n)(x^n) converges absolutely.
K kiddopop New member Joined Sep 14, 2009 Messages 25 Feb 25, 2010 #1 Find all values of x for which the series inf E n=1 (3^n)(x^n) converges absolutely.
G galactus Super Moderator Staff member Joined Sep 28, 2005 Messages 7,216 Feb 26, 2010 #2 ∑n=0∞3nxn\displaystyle \sum_{n=0}^{\infty}\frac{3^{n}}{x^{n}}n=0∑∞xn3n Try the ratio test for absolute convergence. ρ=limn→∞∣uk+1uk∣=limn→∞∣3n+1xn+1⋅xn3n∣=limn→∞∣3x∣=∣3x∣\displaystyle {\rho}=\lim_{n\to {\infty}}\left|\frac{u_{k+1}}{u_{k}}\right|=\lim_{n\to {\infty}}\left|\frac{3^{n+1}}{x^{n+1}}\cdot\frac{x^{n}}{3^{n}}\right|=\lim_{n\to \infty}\left|\frac{3}{x}\right|=\left|\frac{3}{x}\right|ρ=n→∞lim∣∣∣∣∣ukuk+1∣∣∣∣∣=n→∞lim∣∣∣∣∣xn+13n+1⋅3nxn∣∣∣∣∣=n→∞lim∣∣∣∣∣x3∣∣∣∣∣=∣∣∣∣∣x3∣∣∣∣∣ The ratio test for absolute convergence implies that the series converges absolutely if ρ=∣3x∣<1\displaystyle {\rho}=\left|\frac{3}{x}\right|<1ρ=∣∣∣∣∣x3∣∣∣∣∣<1 and diverges if ρ=∣3x∣>1\displaystyle {\rho}=\left|\frac{3}{x}\right|>1ρ=∣∣∣∣∣x3∣∣∣∣∣>1 Thus, the series converges if (−∞, −3)∪(3, ∞)\displaystyle (-\infty, \;\ -3)\cup (3, \;\ \infty)(−∞, −3)∪(3, ∞)
∑n=0∞3nxn\displaystyle \sum_{n=0}^{\infty}\frac{3^{n}}{x^{n}}n=0∑∞xn3n Try the ratio test for absolute convergence. ρ=limn→∞∣uk+1uk∣=limn→∞∣3n+1xn+1⋅xn3n∣=limn→∞∣3x∣=∣3x∣\displaystyle {\rho}=\lim_{n\to {\infty}}\left|\frac{u_{k+1}}{u_{k}}\right|=\lim_{n\to {\infty}}\left|\frac{3^{n+1}}{x^{n+1}}\cdot\frac{x^{n}}{3^{n}}\right|=\lim_{n\to \infty}\left|\frac{3}{x}\right|=\left|\frac{3}{x}\right|ρ=n→∞lim∣∣∣∣∣ukuk+1∣∣∣∣∣=n→∞lim∣∣∣∣∣xn+13n+1⋅3nxn∣∣∣∣∣=n→∞lim∣∣∣∣∣x3∣∣∣∣∣=∣∣∣∣∣x3∣∣∣∣∣ The ratio test for absolute convergence implies that the series converges absolutely if ρ=∣3x∣<1\displaystyle {\rho}=\left|\frac{3}{x}\right|<1ρ=∣∣∣∣∣x3∣∣∣∣∣<1 and diverges if ρ=∣3x∣>1\displaystyle {\rho}=\left|\frac{3}{x}\right|>1ρ=∣∣∣∣∣x3∣∣∣∣∣>1 Thus, the series converges if (−∞, −3)∪(3, ∞)\displaystyle (-\infty, \;\ -3)\cup (3, \;\ \infty)(−∞, −3)∪(3, ∞)
D daon Senior Member Joined Jan 27, 2006 Messages 1,284 Feb 26, 2010 #3 In general, for geometric series, ∑can±kxn=ca±k∑(ax)n\displaystyle \sum \frac{ca^{n \pm k}}{x^n} = ca^{\pm k}\sum \left (\frac{a}{x} \right )^n∑xncan±k=ca±k∑(xa)n converges for ∣x∣>a\displaystyle |x| > a∣x∣>a and ∑can±kxn\displaystyle \sum ca^{n \pm k}x^n∑can±kxn converges for ∣x∣<1a\displaystyle |x|<\frac{1}{a}∣x∣<a1
In general, for geometric series, ∑can±kxn=ca±k∑(ax)n\displaystyle \sum \frac{ca^{n \pm k}}{x^n} = ca^{\pm k}\sum \left (\frac{a}{x} \right )^n∑xncan±k=ca±k∑(xa)n converges for ∣x∣>a\displaystyle |x| > a∣x∣>a and ∑can±kxn\displaystyle \sum ca^{n \pm k}x^n∑can±kxn converges for ∣x∣<1a\displaystyle |x|<\frac{1}{a}∣x∣<a1
