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Show that if $a_n+a_{n+1} \to \alpha$ and $a_n+a_{n+2} \to \beta$ then $\alpha=\beta$ and $a_n \to \alpha/2$.

My textbook denotes $\alpha_n=a_n+a_{n+1}$ and $\beta_n=a_n+a_{n+2}$, notices that $a_n=\frac{1}{2}(\alpha_n+\beta_n-\alpha_{n+1})$ and concludes the proof.

However, I did not see that and I've done something longer using the definition of limit. Can someone check if my work is correct too, please?

My work: let $\epsilon>0$. From the hypotheses on the limits, there exist $N_1,N_2 \in \mathbb{N}$ such that $n \ge N_1$ implies $-\epsilon/2<a_n+a_{n+1}-\alpha<\epsilon/2$ and $n \ge N_2$ implies $-\epsilon/2<a_n+a_{n+2}-\beta<\epsilon/2 \iff -\epsilon/2<-a_n-a_{n+2}+\beta<\epsilon/2$. So, if $n\ge \max\{N_1,N_2\}$, adding the inequalities it is $|a_{n+1}-a_{n+2}-(\alpha-\beta)|<\epsilon$; since this holds for any $\epsilon>0$, it is $|a_{n+1}-a_{n+2}-(\alpha-\beta)|<\epsilon$ and so $a_{n+1}-a_{n+2}\to \alpha-\beta$.

So, using all the three limits definition with $\epsilon/3$, if $n$ is greater than the maximum among the three indexes of these limits hypotheses it is $$|\alpha-\beta|=|a_n-a_n+a_{n+1}-a_{n+1}+a_{n+2}-a_{n+2}+\alpha-\beta|$$ $$\le|a_n+a_{n+1}-\alpha|+|a_n+a_{n+2}-\beta|+|a_{n+1}-a_{n+2}|<\epsilon/3+\epsilon/3+\epsilon/3=\epsilon$$ Hence $\alpha=\beta$ because $\epsilon>0$ is arbitrary.

To show that $a_n \to \alpha/2$, I noticed that since $n+1>n$ the hypothesis $a_n+a_{n+1} \to \alpha$ works with $n+1$ in place of $n$ as well; that is, $a_{n+1}+a_{n+2} \to \alpha$ as well. Hence, using that $\alpha=\beta$, if $n$ is greater than the maximum index among the indexes of the limits hypotheses used with $2\epsilon/3$ it is: $$|a_n-\alpha/2|=\frac{1}{2}|2a_n-\alpha|=\frac{1}{2}|a_n+a_n+a_{n+1}-a_{n+1}+a_{n+2}-a_{n+2}-\alpha+\beta-\beta|$$ $$\le \frac{1}{2}(|a_n+a_{n+1}-\alpha|+|a_n+a_{n+2}-\beta|+|a_{n+1}+a_{n+2}-\alpha|)$$ $$\le \frac{1}{2}\left(\frac{2}{3}\epsilon+\frac{2}{3}\epsilon+\frac{2}{3}\epsilon\right)=\epsilon$$ That is $a_n \to \alpha/2$ because $\epsilon>0$ is arbitrary.

Is this correct? In particular, I am not sure when I add the inequalities and when I say that $a_{n+1}+a_{n+2} \to \alpha$ as well.

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Your proof of the second part is correct, that $a_n\to\alpha/2$ given that $\alpha=\beta$. However your proof that $\alpha=\beta$ is suspect - you invoke the inequality: $$|a_{n+1}-a_{n+2}|\lt\varepsilon/3$$But this is simply false a priori (it is in fact true for large $n$, but only once you have already shown $a_n$ is convergent - it is illegal proof to suppose this inequality beforehand). You have $|a_{n+1}-a_{n+2}-(\alpha-\beta)|\lt\varepsilon/3$ for large $n$, but you cannot forget about the $(\alpha-\beta)$ term.

Adding inequalities is ok (if you're careful...) by the triangle inequality, don't worry, and $a_n+a_{n+1}\to\alpha$ is equivalent to $a_{n+1}+a_{n+2}\to\alpha$ since "for all $n\ge N$" also allows "for all $n+1\ge N+1$"...

To show that $\alpha=\beta$ and complete your proof without error, note that $(a_n+a_{n+1})+(a_{n+2}+a_{n+3})\to2\alpha$ but this is the same as $(a_n+a_{n+2})+(a_{n+1}+a_{n+3})\to2\beta$ hence $\alpha=\beta$.

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  • $\begingroup$ Thank you for correcting my mistake in the first part and checking the whole proof:) sorry if this seems out of context, but lately I am a bit discorauged when I do mistakes on relatively simple problems like this one. Do you think that someone can become a professional mathematician even if he does naive mistakes like these? You can be honest, I prefer direct answers, thanks for your time:) $\endgroup$
    – Gwyn
    Commented Jun 21, 2022 at 16:03
  • $\begingroup$ @Gwyn Everyone makes mistakes; the greatest mathematicians in history have made mistakes. It’s fine. I can’t comment with the authority of a professional mm myself - I am just a student too - but most professionals these days make it through the education system, and that means they made educational mistakes along the way... I too can get very discouraged. But? One year of discouraging mistakes and hours burnt on “simple” exercises and I am today much better than I used to be. Keep going! $\endgroup$
    – FShrike
    Commented Jun 21, 2022 at 17:59
  • $\begingroup$ Thank you for point of view and your kind words. I wish the best for you too! I will take just one last bit of your time, I think I have corrected the first part my proof: $|\alpha-\beta|=\frac{1}{2}|2\alpha-2\beta|=\frac{1}{2}|a_n-a_n+a_{n+1}-a_{n+1}+a_{n+2}-a_{n+2}+\alpha+\alpha-\beta-\beta|\le\frac{1}{2}(|-a_n-a_{n+1}+\alpha|+|a_n+a_{n+2}-\beta|+|a_{n+1}-a_{n+2}-(\alpha-\beta)|)\le\frac{1}{2}(2\epsilon/3+2\epsilon/3+2\epsilon/3)=\epsilon$, having now correctly used the first part of my proof where I show that $a_{n+1}-a_{n+2} \to \alpha-\beta$. Do you confirm it's correct?:) $\endgroup$
    – Gwyn
    Commented Jun 22, 2022 at 3:22
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    $\begingroup$ @Gwyn That is fine now $\endgroup$
    – FShrike
    Commented Jun 22, 2022 at 5:46

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