I would like to know the limit of the sequence: $u_n = (3+\sqrt{5} )^n -(3-\sqrt{5} )^n$.
I tried to involve the binomial theorem but didn't find the answer
limit of a sequence $u_n = (3+\sqrt{5} )^n -(3-\sqrt{5} )^n$.
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sequences-and-series
limits
binomial-theorem
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4It's $\infty$ as second term goes to $0$. – 2017-02-24
2 Answers
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We have $$ u_n = \underbrace{a^n}_{\to\infty} - \underbrace{b^n}_{\to 0} $$ with $a > 1$ and $0 < b < 1$, so $$ \lim_{n\to\infty} u_n = \infty $$
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0You need to write $a>1$ instead of $|a|>1$. – 2017-02-24
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$a>1>b>0 \implies a^n-b^n \to \infty$ as $n \to \infty$ ,since $b^n \to 0$
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0This shows that limit if exists must be $0$. – 2017-02-24
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0@ParamanandSingh Thanks, I made an unfortunate approximation error. – 2017-02-24
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0Your recurrence shows that if $a_{n} $ tends to a limit then the limit must be $0$. But this is not possible as $a_{n} \geq 1$. So $a_n$ does not have a finite limit. But to deduce that it tends to $\infty$ we must show that it is increasing. The second part of your answer (using $a, b$) is correct and depends on the fact that $a^{n} \to \infty$ if $a>1$. In fact the answer depends only on this fact. – 2017-02-24