Does the following series converge conditionally? The series from n=1 to infinity cos(n)/(2^n)
K kiddopop New member Joined Sep 14, 2009 Messages 25 Mar 23, 2010 #1 Does the following series converge conditionally? The series from n=1 to infinity cos(n)/(2^n)
B BigGlenntheHeavy Senior Member Joined Mar 8, 2009 Messages 1,577 Mar 23, 2010 #2 Comparison Test: 1 ≤ n < ∞\displaystyle Comparison \ Test: \ \ 1 \ \le \ n \ < \ \inftyComparison Test: 1 ≤ n < ∞ ∣cos(n)2n∣ ≤ 12n, hence cos(n)2n converges absolutely.\displaystyle \bigg|\frac{cos(n)}{2^{n}}\bigg| \ \le \ \frac{1}{2^{n}}, \ hence \ \frac{cos(n)}{2^{n}} \ converges \ absolutely.∣∣∣∣∣2ncos(n)∣∣∣∣∣ ≤ 2n1, hence 2ncos(n) converges absolutely. Note: This is not an alternating series.\displaystyle Note: \ This \ is \ not \ an \ alternating \ series.Note: This is not an alternating series.
Comparison Test: 1 ≤ n < ∞\displaystyle Comparison \ Test: \ \ 1 \ \le \ n \ < \ \inftyComparison Test: 1 ≤ n < ∞ ∣cos(n)2n∣ ≤ 12n, hence cos(n)2n converges absolutely.\displaystyle \bigg|\frac{cos(n)}{2^{n}}\bigg| \ \le \ \frac{1}{2^{n}}, \ hence \ \frac{cos(n)}{2^{n}} \ converges \ absolutely.∣∣∣∣∣2ncos(n)∣∣∣∣∣ ≤ 2n1, hence 2ncos(n) converges absolutely. Note: This is not an alternating series.\displaystyle Note: \ This \ is \ not \ an \ alternating \ series.Note: This is not an alternating series.
