Evalute $\lim\sup a_n=\left[1-(-2)^n\right]$ I am trying to evaluate the limit $$\lim\sup\limits_{n\to \infty}\sqrt[n]{|a_n|}$$
$$a_n=\left[1-(-2)^n\right]$$.
What I did here is notice that this sequence can be written as the following:$$ a_{2n+1}=3$$ $$a_{2n}=1$$
From here it is really clear that the $\lim\sup a_n=3$, but this is a wrong answer, the correct answer is $2$.
I do not understand what is wrong in my assumptions
 A: Your computations are false. I am guessing that you confused $(-2)^n$ with $2 \times(-1)^n$.
The answer is $2$ indeed :
You can check that $|a_n| = 2^n + e_n$, where $(e_n)$ is bounded. Thus, $|a_n|^{\frac1n} = \exp(\frac1n\ln (2^n + e_n)) = \exp\left(\frac{\ln(2^n)+\ln\left(1+\frac{e_n}{2^n}\right)}n\right)\to 2$.
A: You're computing the values wrongly:
$$
a_n=\begin{cases}
1-2^n & \text{$n$ even} \\[6px]
1+2^n & \text{$n$ odd}
\end{cases}
$$
Thus
$$
b_n=\sqrt[n]{|a_n|}=\begin{cases}
\sqrt[n]{2^n-1} & \text{$n$ even} \\[6px]
\sqrt[n]{2^n+1} & \text{$n$ odd}
\end{cases}
$$
Thus you actually have
$$
\lim_{n\to\infty} b_n=2
$$
in particular the superior limit is $2$. It's not difficult to show that if the even and the odd terms of a sequence converge to the same limit, then also the whole sequence converges to that limit. In this case it's even simpler:
$$
b_n=2\sqrt[n]{1-\dfrac{(-1)^n}{2^n}}
$$
A: As we are looking for $\lim\sup\limits_{n\to \infty}\sqrt[n]{|a_n|}$, then may by most  easy is to consider sub sequence $\sqrt[2n+1]{|a_{2n+1}|} =\sqrt[2n+1]{2^{2n+1}+1 }=2 \cdot \sqrt[2n+1]{1+\frac{1}{2^{2n+1}}}$ and then show $\lim_{}\sqrt[2n+1]{1+\frac{1}{2^{2n+1}}}=1$:
$$1<\sqrt[2n+1]{1+\frac{1}{2^{2n+1}}}<\sqrt[n]{2} \to 1$$
For even sub sequense we have $\sqrt[2n]{|a_{2n}|}=\sqrt[2n]{2^{2n}-1}=2 \cdot \sqrt[2n]{1-\frac{1}{2^{2n}}} \to2$
