I want to know radius of convergence. $\sum ^{\infty }_{n=0}\left\{ 3+\left( -1\right) ^{n}\right\} x^{n}$

Question

$$\sum ^{\infty }_{n=0}\left\{ 3+\left( -1\right) ^{n}\right\} x^{n}$$ I used ratio test to find radius of convergence.

$$\lim _{n\rightarrow \infty }\left| \dfrac {\left\{ 3+\left( -1\right) ^{n+1}\right\} x^{n+1}}{\left\{ 3+\left( -1\right) ^{n}\right\} x^{n}}\right|=\lim _{n\rightarrow \infty }\left| \dfrac {\left\{ 3+\left( -1\right) ^{n+1}\right\} x}{\left\{ 3+\left( -1\right) ^{n}\right\} }\right|$$ But I don't know what to do next. Please tell me how to solve.

Answer 1

Hint. In this case you should use the root test (the general form is with $\limsup$): $$R=\frac{1}{\limsup_{n\to\infty}\sqrt[n]{|a_n|}}.$$ Now note that $$a_n=\begin{cases} 4&\text{ if $n$ is even,}\\ 2&\text{ if $n$ is odd.} \end{cases}$$ Can you take it from here?

Answer 2

The limsup ratio test will give you a bound for the radius of convergence. To actually compute this ratio precisely, we need a different method. Consider the following (suspect) equality: $$\sum_{n=0}^\infty (3 + (-1)^n)x^n = \sum_{n=0}^\infty 3x^n + \sum_{n=0}^\infty(-1)^nx^n.$$ This holds true if both sums on the right hand side converge simultaneously. Both of these series are geometric, with common ratios $x$ and $-x$ respectively. They converge precisely when $|x| = |-x| < 1$. So, if $|x| < 1$, then the series converges, and hence the radius of convergene is at least $1$.

Now, we need to establish whether the series converges for $|x| \ge 1$. Unfortunately, both of the series on the right hand side diverging does not imply that their sum diverges. What we can do is show that the terms $(3 + (-1)^n)x^n$ do not converge to $0$, which will mean the series on the left fails the divergence test.

If $|x| \ge 1$, then $$|(3 + (-1)^n)x^n| = (3 + (-1)^n)|x|^n \ge 2|x|^n \ge 2,$$ hence $(3 + (-1)^n)x^n$ cannot converge to $0$ (if it did, then $|(3 + (-1)^n)x^n|$ would have to become arbitrarily small). Thus the radius is indeed $1$, and the interval of convergence is $(-1, 1)$.

Answer 3

The root test also works here. Note $$ \lim_{n\to\infty} 4^{1/n} = 1\qquad\text{and}\qquad\lim_{n\to\infty} 2^{1/n} = 1 $$ so $$ \lim_{n\to\infty} (3+(-1)^n)^{1/n} = 1 $$ so that the radius of convergence is $1$.