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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

FiMePr
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Sagigever
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3 Answers3

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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, $|a_n|^{\frac1n} = \exp(\frac1n\ln (2^n + e_n)) = \exp\left(\frac{\ln(2^n)+\ln\left(1+\frac{e_n}{2^n}\right)}n\right)\to 2$.

Matcha Latte
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FiMePr
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    Indeed, it does not. I am guessing that this is the mistake the OP made. – FiMePr Jun 28 '20 at 08:56
  • By $|a_n|$, I mean the modulus of $a_n$. It is not hard to see that $a_n$ alternates signs. But its modulus is roughly $2^n$ – FiMePr Jun 28 '20 at 08:59
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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: $$ b_n=2\sqrt[n]{1-\dfrac{(-1)^n}{2^n}} $$

egreg
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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 $\sqrt[2n]{|a_{2n}|}=\sqrt[2n]{2^{2n}-1}=2 \cdot \sqrt[2n]{1-\frac{1}{2^{2n}}} \to2$

zkutch
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