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Let $(a_n)$ be a strictly increasing sequence of real numbers such that $a_0\geq 1$ and $a_n\rightarrow \infty$ as $n\rightarrow \infty.$

The following limit is always convergent or which condition may be added on $(a_n)$ the given limit to be convergent? $$\lim_{n\to\infty}\left(1-\frac{1}{a_n}\right)^n$$

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Hint : Write the limit as $$\left[\left(1 - \frac{1}{a_n}\right)^{a_n} \right]^{\frac{n}{a_n}}.$$ The limit inside the bracket is $\frac{1}{e}$; what must the value of $n/{a_n}$, as $n\to \infty$, be, to ensure that the limit converges?

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Since $y = \mathrm{e}^x$ is a continuous function and$$ \left(1 - \frac{1}{a_n}\right)^n = \exp\left(n \ln\left(1 - \frac{1}{a_n}\right)\right), $$ then$$ \lim_{n \to \infty} \left(1 - \frac{1}{a_n}\right)^n \ \text{exists} \Longleftrightarrow \lim_{n \to \infty} n \ln\left(1 - \frac{1}{a_n}\right) \ \text{exists}. $$ Given that $a_n \to \infty \ (n \to \infty)$, thus$$ \lim_{n \to \infty} \frac{1}{a_n} = 0 \Longrightarrow \ln\left(1 - \frac{1}{a_n}\right) \sim -\frac{1}{a_n} \ (n \to \infty). $$ Therefore,$$ \lim_{n \to \infty} n \ln\left(1 - \frac{1}{a_n}\right) \ \text{exists} \Longleftrightarrow \lim_{n \to \infty} \frac{n}{a_n} \ \text{exists}. $$

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  • $\begingroup$ Condition on $a_n$ is not sufficient for $\lim_{n\to\infty} n/a_n$ to be convergent? $\endgroup$ – Raio Jan 8 '18 at 11:26
  • $\begingroup$ What condition exactly is it? $\endgroup$ – Saad Jan 8 '18 at 11:29
  • $\begingroup$ Some small comments, it is important that $e^x$ is injective. For instance $a_n^2$ can exist and $a_n$ may not exist. Also, when you say $$ \lim_{n \to \infty} \left(1 - \frac{1}{a_n}\right)^n \ \text{exists} \Longleftrightarrow \lim_{n \to \infty} n \ln\left(1 - \frac{1}{a_n}\right) \ \text{exists}. $$ The left limit converges to $0$ when the right limit converges to $-\infty$, and the left limit converges to $\infty$ when the right limit converges to $\infty$ $\endgroup$ – clark Jan 8 '18 at 11:29
  • $\begingroup$ $(a_n)$ is strictly increasing and $a_0\geq 1$ and $a_n\rightarrow \infty$ as $n\rightarrow \infty.$ $\endgroup$ – Raio Jan 8 '18 at 11:31
  • $\begingroup$ @andii That's not sufficient and there're counterexamples. $\endgroup$ – Saad Jan 8 '18 at 11:34

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