Let $z_0=3$ and $ z_{n+1}=\dfrac{2}{z_n +1} $ and $t_n=\dfrac{z_n-1}{z_n+2}$ prove that $t_n$ is a geometric sequence find its general form.
A geometric sequence
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real-analysis
sequences-and-series
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0Do you substitute $t_n$, $t_{n+1}$ to delete $z$'s from general term? – 2017-01-08
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0Help me @MyGlasses – 2017-01-08
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0Try $$\frac{t_{n+1}}{t_n}$$. – 2017-01-08
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0I've found nothing – 2017-01-08
2 Answers
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$$ \frac{t_{n+1}}{t_{n}}=t_{n+1}\frac{1}{t_{n}}= \frac{z_{n+1}-1}{z_{n+1}+2}\frac{z_{n}+2}{z_{n}-1}= \frac{\frac{2}{z_{n}+1}-1}{\frac{2}{z_{n}+1}+2}\frac{z_{n}+2}{z_{n}-1}= \frac{-(z_{n}-1)}{2(z_{n}+2)}\times\frac{z_{n}+2}{z_{n}-1}= -\frac12 $$ which is a geometric sequence $\displaystyle t_0=\frac25$.
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Try harder:
$$\frac{t_{n+1}}{t_n}=\frac{z_{n+1}-1}{z_{n+1}+2}\cdot\frac{z_n+2}{z_n-1}=\frac{\frac2{z_n+1}-1}{\frac2{z_n+1}+2}\cdot\frac{z_n+2}{z_n-1}=$$
$$=\frac{1-z_n}{2(z_n+2)}\cdot\frac{z_n+2}{z_n-1}=\ldots$$