Finding Radius of Convergence for Series: n/2^n and 1/(4+(-1)^n)^3n

[mathmari]

Hey!

I want to find for the following series the radius of convergence and the set of $x\in \mathbb{R}$ in which the series converges.

1. $\displaystyle{\sum_{n=0}^{\infty}\frac{n}{2^n}x^{n^2}}$
2. $\displaystyle{\sum_{n=0}^{\infty}\frac{1}{(4+(-1)^n)^{3n}}(x-1)^{3n}}$

I have done the following:

1. $a_n=\frac{n}{2^n}$
   $$|\frac{a_n}{a_{n+1}}|=\frac{n2^{n+1}}{(n+1)2^n}=\frac{2n}{n+1}=\frac{2}{1+\frac{1}{n}}$$
   So, $$R=\lim_{n\rightarrow \infty}|\frac{a_n}{a_{n+1}}|=\lim_{n\rightarrow \infty}\frac{2}{1+\frac{1}{n}}=2$$
   The radius of convergence is $2$. The series converges at $|x|<2$ and diverges at $|x|>2$.
   For $x=\pm 2$ we have the following:
   $$\sum_{n=0}^{\infty}\frac{n}{2^n}(\pm 2)^{n^2}$$
   How could we continue?
   
2. $a_n=\frac{1}{(4+(-1)^n)^{3n}}$
   $$\sqrt[n]{|a_n|}=\sqrt[n]{\frac{1}{(4+(-1)^n)^{3n}}}=\frac{1}{(4+(-1)^n)^3}$$
   For $n=2k$ : $$\lim_{k\rightarrow \infty}\frac{1}{(4+(-1)^{2k})^3}=\frac{1}{5^3}$$
   For $n=2k+1$ : $$\lim_{k\rightarrow \infty}\frac{1}{(4+(-1)^{2k+1})^3}=\frac{1}{3^3}$$
   So, $\frac{1}{R}=\lim\sup\sqrt[n]{|a_n|}=\frac{1}{3^3}$.
   The radius of convergence is $3^3=27$. The series converges at $|x|<27$ and diverges at $|x|>27$.
   For $x=27$ we have the following:
   $$\sum_{n=0}^{\infty}\frac{1}{(4+(-1)^n)^{3n}26^{3n}}$$ How could we check if the series converges? (Wondering)
   For $x=-27$ we have the following:
   $$\sum_{n=0}^{\infty}\frac{1}{(4+(-1)^n)^{3n}(-28)^{3n}}=\sum_{n=0}^{\infty}\frac{1}{(4+(-1)^n)^{3n}28^{3n}(-1)^n}$$ Is this an alternating series? Do we use here the Leibniz criterium? (Wondering)

 

[Prove It]

You have the ratio test wrong. It's actually $\displaystyle \begin{align*} \lim_{n \to \infty} \left| \frac{a_{n+1}}{a_n} \right| \end{align*}$ that you need to evaluate.

 

[mathmari]

You have the ratio test wrong. It's actually $\displaystyle \begin{align*} \lim_{n \to \infty} \left| \frac{a_{n+1}}{a_n} \right| \end{align*}$ that you need to evaluate.

Then we would have $\displaystyle \begin{align*} \frac{1}{R}=\lim_{n \to \infty} \left| \frac{a_{n+1}}{a_n} \right| \end{align*}$ or not? (Wondering)

 

[Opalg]

I want to find for the following series the radius of convergence and the set of $x\in \mathbb{R}$ in which the series converges.

1. $\displaystyle{\sum_{n=0}^{\infty}\frac{n}{2^n}x^{n^2}}$
2. $\displaystyle{\sum_{n=0}^{\infty}\frac{1}{(4+(-1)^n)^{3n}}(x-1)^{3n}}$

For a simple power series $\sum a_nx^n$ the radius of convergence is given by $\frac1R = \lim_{n\to\infty}\Bigl|\frac{a_{n+1}}{a_n}\Bigr|$ (provided that limit exists). But these are not simple power series, because the power of $x$ is not $n$, but $n^2$ in 1., and $3n$ in 2. In 2., there is the additional complication that the variable is not $x$, but $x-1$.

So to answer these questions you need to go back to the more general form of the ratio test, which says that a series converges if the limit as $n\to\infty$ of the ratio of the $(n+1)$th term to the $n$th term is less than $1$, and it diverges if that limit is greater than $1$.

For problem 1., that ratio is $\left|\dfrac{\frac{(n+1)x^{(n+1)^2}}{2^{n+1}}}{\frac{nx^{n^2}}{2^n}}\right| = \Bigl|\dfrac{(n+1)x^{2n+1}}{2n}\Bigr|.$

For problem 2., you would do best to look at the even-numbered and odd-numbered terms separately. If $n$ is even then the $n$th term is $\Bigl(\dfrac{x-1}5\Bigr)^{3n}.$ If $n$ is odd then it is $\Bigl(\dfrac{x-1}3\Bigr)^{3n}.$