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If the sequence converges to $L$ then:

$lim \frac{1}{x_{n+2}} = lim \frac{1}{x_{n+1}} + lim \frac{1}{x_n} \implies \frac{1}{L} = \frac{1}{L} + \frac{1}{L} \implies L = 2L \implies L = 0$.

So I know that if the sequence converges then she will converge to $0$. How can I prove that the sequence converges? (It's clearly bounded and monotone, but how can I prove that formally?)

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    $\begingroup$ If $L=0$ then $\lim\frac1{x_n}$ doesn't exist $\endgroup$ – Kenny Lau Jun 15 '18 at 2:34
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    $\begingroup$ It's clearly bounded $x_n \gt 0\,$. and monotone $\frac{1}{x_{n+2}}=\frac{1}{x_{n+1}}+\frac{1}{x_n} \gt \frac{1}{x_{n+1}}\,$. $\endgroup$ – dxiv Jun 15 '18 at 2:38
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Claim: $0<x_n \leq 1/n$ for all $n\geq 2$.

Note that $x_2 = 1/2$ and $x_3 = 1/3$ so the base cases hold. Suppose that it is true up to some $x_n$. Then $$ \frac{1}{x_{n+1}} = \frac{1}{x_n}+\frac{1}{x_{n-1}}\geq 2n-1 \geq n+1, $$ where the last inequality follows since $n\geq 2$. Thus $x_{n+1}\leq \frac{1}{n+1}$ and by induction the claim follows.

Now, by the squeeze theorem it is clear that $x_n\to 0$.

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Let $y_n = \dfrac1{x_n}$.

Then, $y_0 = 1$ and $y_1 = 1$ and $y_{n+2} = y_{n+1} + y_n$, so $y_n$ is just Fibonacci, and it is well-known that Fibonacci goes to infinity, so $x_n = \dfrac1{y_n}$ goes to $0$.

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  • $\begingroup$ It's enough to notice that $y_{n+2}\ge y_{n+1}+1$ , hence $y_{n+2}\ge n$, hence $y_n \to +\infty$ $\endgroup$ – leonbloy Jun 15 '18 at 3:06

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