I read somewhere that $$\frac{\big(\frac{3}{2}\big)^{99}-1}{\big(\frac{3}{2}\big)^{100}-1}\approx\frac{1}{\big(\frac{3}{2}\big)}$$I don't know how to have it. Please let me know how this is approximated.
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$(3/2)^{99}$ and $(3/2)^{100}$ are both really big compared to the $-1$ term in the numerator and the denominator. So the idea is that you can ignore the $-1$ terms and evaluate the fraction as $(3/2)^{99} / (3/2)^{100} = 1/(3/2)$.
Aaron Golden
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So quick! Thanks. – Silent Oct 07 '13 at 07:16
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$+1$ for your post. – Mikasa Oct 07 '13 at 15:14
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Another point of view: $$(3/2)\left((3/2)^{99}-1\right)=(3/2)^{100}-(3/2)=(3/2)^{100}-1-0.5$$
Mikasa
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And the numerator is just $0.5$ less than the denominator, so we get approximation, right? – Silent Oct 07 '13 at 07:27
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I don't understand how this clarifies the approximation in the question. @BabakS, could you expand this a bit? – Aaron Golden Oct 07 '13 at 08:07
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@AaronGolden: Personally, I saw your post more descriptive and I posted just an small hint. I don't know why the OP chose mine, however!. – Mikasa Oct 07 '13 at 15:13
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Let $x>1$ and $q_n = \frac{x^n-1}{x^{n+1}-1}$.
We see that $q_n = \frac{1-\frac{1}{x^n}}{x-\frac{1}{x^n}}$, from which we have $\lim_n q_n = \frac{1}{x}$.
In this case we have $x= \frac{3}{2}$, hence $\lim_n \frac{(\frac{3}{2})^n-1}{(\frac{3}{2})^{n+1}-1} = \frac{1}{(\frac{3}{2})}$.
$n=99$ is sufficiently 'large' that the approximation works.
copper.hat
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