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We define the sequences $a_n$ and $b_n$ as $(a_n+b_n\sqrt2)=(1+\sqrt2)^n$. The exercise has those:
a) Calculate the limit as n tends to infinity of $a_n/b_n$.
b) Calculate $S_1=\sum_{i=1}^{n}(1+\sqrt2)^i$, $S_2=\sum_{i=1}^{n}a_i$, $S_3=\sum_{i=1}^{n}b_i$ for any $n\in \mathbb{N}$ and find the limit as $n$ tends to infinity of $\frac{S_2}{S_3}$
c) Prove that there are unique $c,d\in\mathbb{Z}$ such that $a_{n+2}= ca_{n+1} +da_{n}$ and $b_{n+2}= cb_{n+1} +db_{n}$ for any $n\in \mathbb{N}$.

I am having a lot of trouble dealing with those, some help?

Arturo Magidin
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MiguelCG
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    Please edit the question to limit it to a specific problem with enough detail to identify an adequate answer. – Community Nov 14 '23 at 15:13
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    Hint : multiplication by $1 + \sqrt{2}$ is a linear transformation on $\mathbb{Q}(\sqrt(2))$ as a vector space over $\mathbb{Q}$. Find its matrix representation. – David Lui Nov 14 '23 at 15:18
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    A closely related thread. Many of the same calculations have been carried out there, but not all. I'm not gonna call it a duplicate, as there are differences in focus. Would not be kosher for me to vote that way anyway, as I happened to answer the other one. However, see Zev Chonoles's answer for the recurrence formulas. – Jyrki Lahtonen Nov 14 '23 at 15:45
  • Thanks, for the comments!! For the last part of the exercise I saw really usefull the comment of Sawarnik on the thread u linked. – MiguelCG Nov 14 '23 at 16:39

1 Answers1

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Since $\{a_n\},\{b_n\}$ are increasing, you can prove that $$\lim_{n\to\infty}a_n=\lim_{n\to\infty}b_n=\infty. $$ From $a_n^2-2b_n^2=(-1)^n$, you have $$ \bigg(\frac{a_n}{b_n}\bigg)^2=2-\frac{(-1)^n}{b_n^2}$$ which implies $$ \lim_{n\to\infty}\bigg(\frac{a_n}{b_n}\bigg)^2=2$$ or $$ \lim_{n\to\infty}\frac{a_n}{b_n}=\sqrt2.$$ So c) is done.

Clearly $$ S_1=\sum_{i=1}^n(a_i+b_i\sqrt2)=S_2+S_3\sqrt2=\frac{\sqrt2}2(1+\sqrt2)((1+\sqrt2)^n-1).\tag1$$ It is easy to see $$ S_2-S_3\sqrt2=-\frac{\sqrt2}2(1-\sqrt2)((1-\sqrt2)^n-1).\tag2$$ From (1) and (2), you have \begin{eqnarray} S_2=\frac{\sqrt2}4\bigg[(1+\sqrt2)((1+\sqrt2)^n-1)-(1-\sqrt2)((1-\sqrt2)^n-1)\bigg]\\ S_3=\frac{1}4\bigg[(1+\sqrt2)((1+\sqrt2)^n-1)+(1-\sqrt2)((1-\sqrt2)^n-1)\bigg] \end{eqnarray} and hence $$ \lim_{n\to\infty}\frac{S_2}{S_3}=\lim_{n\to\infty}\frac{\sqrt2\bigg[(1+\sqrt2)((1+\sqrt2)^n-1)-(1-\sqrt2)((1-\sqrt2)^n-1)\bigg]}{(1+\sqrt2)((1+\sqrt2)^n-1)+(1-\sqrt2)((1-\sqrt2)^n-1)}=\sqrt2. $$ So d) is done. It should be easy to get e).

MiguelCG
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xpaul
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  • Great answer, thank you so much. However, am I mistaking something or the expressions of $S_2$ and $S_3$ should be $S_2=\frac{\sqrt2}4\bigg[(1+\sqrt2)((1+\sqrt2)^n-1)-(1-\sqrt2)((1-\sqrt2)^n-1)\bigg]$ and $S_3=\frac{1}4\bigg[(1+\sqrt2)((1+\sqrt2)^n-1)+(1-\sqrt2)((1-\sqrt2)^n-1)\bigg]$. – MiguelCG Nov 14 '23 at 16:08
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    You are right. I changed it. – xpaul Nov 14 '23 at 16:13