Question

what values of x does the power series converge absolutely or converge conditionally ?

Answer 1

i got the x values = 2 and -2

Answer 2

@SkyOfNight @skullpatrol

Answer 3

Let \(f(x)\) be the summation and \(g(x)\) the sum minus the first term:
\[g(x)=\sum_{n=1}^\infty \frac{x^n}{2^nn}\]This converges absolutely by the ratio test for \(|x|<2\), because
\[\lim_{n\to\infty}\left|\frac{c_{n+1}}{c_n}\right|=\lim_{n\to\infty}\left|\frac{x^{n+1}}{2^{n+1}(n+1)}\times\frac{2^nn}{x^n}\right|=\frac{|x|}{2}\]Check the endpoints of this interval of convergence. When \(x=2\), you have
\[g(2)=\sum_{n=1}^\infty \frac{2^n}{2^nn}=\sum_{n=1}^\infty \frac{1}{n}\]but this diverges. Whereas when \(x=-2\), you have
\[g(-2)=\sum_{n=1}^\infty\frac{(-2)^n}{2^nn}=\sum_{n=1}^\infty\frac{(-1)^n}{n}\]You can check (if you don't already know) that this alternating series converges, but not absolutely because
\[\sum_{n=1}^\infty\left|\frac{(-1)^n}{n}\right|=\sum_{n=1}^\infty\frac{1}{n}\to\infty\]