# $a_n+a_{n+1} \to \alpha$ and $a_n+a_{n+2} \to \beta$ then $\alpha=\beta$ and $a_n \to \alpha/2$

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 $$-\epsilon/2 and $$n \ge N_2$$ implies $$-\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.

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 $$\alpha=\beta$$.

• 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:)
– Gwyn
Commented Jun 21, 2022 at 16:03
• @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! Commented Jun 21, 2022 at 17:59
• 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?:)
– Gwyn
Commented Jun 22, 2022 at 3:22
• @Gwyn That is fine now Commented Jun 22, 2022 at 5:46