Could anyone help me with the following problem? I'm getting $1$ as the answer, but I found the solution (without justification) to this problem online and it says the answer is $1/2$.
Determine the radius of convergence of $\sum (1+(-1)^n)z^n$.
Here's my solution:
The radius of convergence, by definition, is given by $1/L$, where $L = \lim\sup |c_n|^{1/n}$. Here, $c_n = 1 + (-1)^n$. Observe that
$$ \lim\sup |1 + (-1)^n|^{1/n} = \lim_{n\to\infty} (1 + (-1)^n)^{1/n} = 1 $$
according to WolframAlpha. Thus, the radius of convergence is $1/1 = 1$.
My guess is that the problem intended for the summation to be $\displaystyle\sum_{n = 0}^{\infty}(1+(-1)^n)^nz^n$.
Since $1+(-1)^n = \begin{cases} 2 & n \ \text{is even}\\ 0 & n \ \text{is odd}\end{cases}$, we have $|c_n|^{1/n} = \left|\left(1+(-1)^n\right)^n\right|^{1/n} = \begin{cases} 2 & n \ \text{is even}\\ 0 & n \ \text{is odd}\end{cases}$.
Thus, $\displaystyle\limsup_{n \to \infty}|c_n|^{1/n} = 2$, and the radius of convergence is $\dfrac{1}{2}$.
However, as the problem is written, $c_n = 1+(-1)^n = \begin{cases} 2 & n \ \text{is even}\\ 0 & n \ \text{is odd}\end{cases}$.
So, the summation is $\displaystyle\sum_{n = 0}^{\infty}(1+(-1)^n)z^n = \sum_{\substack{n = 0 \\ n \ \text{is even}}}^{\infty}2z^n = \sum_{m = 0}^{\infty}2z^{2m}$.
This geometric series converges iff $|x^2| < 1$, i.e. $|x| < 1$, so it has a radius of convergence of $1$.
Hence, your answer is correct for the problem as it is written.