What is $\lim_{n \rightarrow \infty}(1-c^n)^\frac{n}{2}$? Where $c \in [0, 0.5)$, and $n$ is an integer. How do I calculate it?
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1Use $(1+x)^n\sim1+nx$. – PNDas Dec 06 '20 at 15:12
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Does this answer your question (write $n=2m$ and $a=-c^2)$?(https://math.stackexchange.com/questions/35486/proof-that-this-limit-equals-ea) – Integrand Dec 06 '20 at 16:38
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Hint. Let $|c|<1$. One may write $$ (1-c^n)^\frac{n}{2}=e^{\frac{n}{2}\ln\left(1-c^n \right)} $$ noticing that, as $n \to \infty$, $c^n \to 0$ and $$ \frac{\ln\left(1-c^n \right)}{-c^n}\to 1. $$
Olivier Oloa
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Let $\epsilon > 0$. Then for large $n$ we have $$ 0 < c^n < \frac{\epsilon}{n} \\ 1 > (1-c^n) > 1-\frac{\epsilon}{n} \\ 1 > (1-c^n)^{n/2} > \left(1-\frac{\epsilon}{n}\right)^{n/2} $$ Now $\left(1-\frac{\epsilon}{n}\right)^{n/2} \to e^{-\epsilon}$ so $$ 1 \ge \limsup_n(1-c^n)^{n/2} \ge \liminf_n(1-c^n)^{n/2} \ge e^{-\epsilon} $$ Next, $\lim_{\epsilon \to 0} e^{-\epsilon} = 1$, so finally $$ 1 \ge \limsup_n(1-c^n)^{n/2} \ge \liminf_n(1-c^n)^{n/2} \ge 1 $$ and therefore $\lim_n(1-c^n)^{n/2}$ exists and equals $1$.
GEdgar
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