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!