Radius & Interval of Convergence for Power Series

[azatkgz]

Homework Statement

Find the radius and interval of convergence for the following power series.
$$\sum_{n = 2}^{\infty}\frac {(1 + 2cos\frac {\pi n}{4})^n}{lnn}x^n$$

The Attempt at a Solution

$$R = \frac {1}{\lim_{n\rightarrow\infty}\sqrt [n]{\frac {(1 + 2cos\frac {\pi n}{4})^n}{lnn}}} = \lim_{n\rightarrow\infty}\frac {e^{\frac {ln(lnn)}{n}}}{(1 + 2cos\frac {\pi n}{4})}$$
In answers $$R=\frac{1}{3}$$.
$$\lim_{n\rightarrow\infty}e^{\frac {ln(lnn)}{n}} = 1$$.Then is

$$\lim_{n\rightarrow\infty}(1 + 2cos\frac {\pi n}{4}) = 3$$?

As I know usually $$\lim_{x\rightarrow 0}cosx=1$$,not

$$\lim_{x\rightarrow\infty}cosx=1$$

 

[dalle]

try using $$\frac{1}{R} =\limsup_{n \to \infty} \sqrt[n]{| a_n |}$$

 

[HallsofIvy]

Homework Statement

Find the radius and interval of convergence for the following power series.
$$\sum_{n = 2}^{\infty}\frac {(1 + 2cos\frac {\pi n}{4})^n}{lnn}x^n$$

The Attempt at a Solution

$$R = \frac {1}{\lim_{n\rightarrow\infty}\sqrt [n]{\frac {(1 + 2cos\frac {\pi n}{4})^n}{lnn}}} = \lim_{n\rightarrow\infty}\frac {e^{\frac {ln(lnn)}{n}}}{(1 + 2cos\frac {\pi n}{4})}$$
In answers $$R=\frac{1}{3}$$.
$$\lim_{n\rightarrow\infty}e^{\frac {ln(lnn)}{n}} = 1$$.Then is

$$\lim_{n\rightarrow\infty}(1 + 2cos\frac {\pi n}{4}) = 3$$?

As I know usually $$\lim_{x\rightarrow 0}cosx=1$$,not

$$\lim_{x\rightarrow\infty}cosx=1$$

cos(x) itself does not approach ANYTHING as x goes to infinity, it cycles back and forth between -1 and 1. However, at any point at which cos($\pi n/4$) is 1, 2 cos($\pi n/4$) is 2 and 1+ 2cos($\pi n/4$) is 3. That is the largest the denominator can get.

 

[azatkgz]

I see.Thanks!