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Let $a, b >0$ and $a \ll b$, $a<1$. Does $\lim_{n\to \infty} \left(\frac{a}{a+bn}\right)^{\frac{1}{n}}$ exist ?

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  • $\begingroup$ tried anything? what about taking logartihm? $\endgroup$ Jul 11, 2017 at 1:35
  • $\begingroup$ @SiongThyeGoh: Just got it. It is 1. $\endgroup$ Jul 11, 2017 at 1:36

2 Answers 2

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It exists for all real $a$ and $b$. $$\lim_{n\to\infty}\left(\frac{a}{a+bn}\right)^{\frac{1}{n}}=\lim_{n\to\infty}\exp\left(\frac{1}{n}\ln\left(\frac{a}{a+bn}\right)\right)=\lim_{n\to\infty}e^{-\frac{b}{a+bn}}=1$$ where the second equality is from applying L'H$\hat{\text{o}}$pital, gotta take it to the limit hospital.

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Using \begin{align} \frac{a}{a + b \, n} &= 1 - \frac{1}{1 + \frac{a}{b \, n}} \\ &= 1 - \left( 1 - \frac{a}{b \, n} + \frac{a^{2}}{b^{2} \, n^{2}} + \mathcal{O}\left(\frac{1}{n^{3}}\right) \right) \\ &= \frac{a}{b \, n} - \left(\frac{a}{b \, n}\right)^{2} + \mathcal{O}\left(\left(\frac{a}{b \, n}\right)^{3}\right) \\ &= \frac{a}{b \, n} \, \left( 1 - \frac{a}{b \, n} + \mathcal{O}\left(\left(\frac{a}{b \, n}\right)^{2}\right) \right) \end{align} then \begin{align} \lim_{n \to \infty} \left( \frac{a}{a + b \, n} \right)^{\frac{1}{n}} &= \lim_{n \to \infty} \, \left(\frac{a}{b \, n}\right)^{\frac{1}{n}} \, \left( 1 - \frac{a}{b \, n} + \mathcal{O}\left(\left(\frac{a}{b \, n}\right)^{2}\right) \right)^{\frac{1}{n}}. \end{align} By inverting $n$, ie $n \to \frac{1}{n}$, the limit becomes \begin{align} \lim_{n \to \infty} \left( \frac{a}{a + b \, n} \right)^{\frac{1}{n}} &= \lim_{n \to 0} \, \left(\frac{a \, n}{b}\right)^{n} \, \left( 1 - \frac{a \, n}{b} + \mathcal{O}\left(\left(\frac{a \, n}{b}\right)^{2}\right) \right)^{n} = 0^{0} = 1. \end{align}

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