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I would like to prove that $1$ is an upper bound of $a_{n}:=(-1)^n$ using induction. I am stuck in the inductive step, namely, if $1\geq (-1)^n$, then $1\geq (-1)^{n+1}$.

I know that $1\geq (-1)^n \Leftrightarrow (-1)\times 1\geq (-1)\times (-1)^n \Leftrightarrow 1\geq -(-1)^{n+1}$. But that seems to go nowhere.

Edit: typo: it has to be $1\geq (-1)^n \Leftrightarrow (-1)\times 1\leq (-1)\times (-1)^n \Leftrightarrow 1\geq -(-1)^{n+1}$.

pompeu
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1 Answers1

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One way to go is to make your inductive statement be that $a_{n+1} \in \{-1,1\}$, a set that is clearly upper bounded by $1$.

Eric Towers
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  • Thank you, but isn't there a more, let's say, algebraic way of doing it? – pompeu Oct 20 '19 at 22:32
  • @pompeu I don't understand your not liking this solution. We suppose that $a_n$ is either $1$ or $-1$. We show that such an assumption will further imply that $a_{n+1}$ is also either $1$ or $-1$. The proof is simply that if $a_n$ was $1$, then we see that $a_{n+1}=-1\times a_n = -1\times 1 = -1$. In the case that instead $a_n$ was $-1$ we similarly arrive at $a_{n+1}=(-1)\times (-1)=1$, QED – JMoravitz Oct 20 '19 at 22:43
  • @pompeu another way of doing it, is as described, by showing by induction that $a_{2n} = 1$ for all $n$ and similarly that $a_{2n+1}=-1$ for all $n$, noting that $a_{2(n+1)}=a_{2n+2}=(-1)\times a_{2n+1} = (-1)\times (-1)\times a_{2n}$, and similarly for the other case. – JMoravitz Oct 20 '19 at 22:46