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Why is $\lim\limits_{n\to\infty}\displaystyle\Big(1-\frac{\sigma^2\xi^2}{2n}+o(\frac1n)\Big)^n=\large e^{-\frac{\sigma^2\xi^2}{2}}$ ?

Why has $o(\frac{1}{n})$ no effect on the term ? Can I also conclude that $\lim\limits_{n\to\infty}\displaystyle\Big(1+o(\frac1n)\Big)^n=1$ ?

Ayman Hourieh
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OBDA
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1 Answers1

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Look at $$ \bigg(1+\frac{1}{n^{1+ \epsilon}}\bigg)^n = \bigg(1+\frac{1}{n^{1 +\epsilon} } \bigg)^{n^{1+ \epsilon -\epsilon}} = \bigg( 1+\frac{1}{n^{1 +\epsilon} } \bigg)^{n^{1+ \epsilon} \cdot n^{-\epsilon}} \leq e^{n^{-\epsilon}} \to_n 1 $$

EDIT: another way: the lower bound is due to Bernoulli inequality. Since $n > 0$ $$ (1+ o(\frac{1}{n})^n \geq (1+n o(\frac{1}{n})) \to_n 1 $$ upper bound: $$ (1+o(\frac{1}{n}))^n \leq e^{n o(\frac{1}{n})} \to_n 1 $$ this is due to $n o(\frac{1}{n}) \to_n 0$.

Alex
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  • Awesome. I think this is sufficient. – OBDA Jun 10 '14 at 13:27
  • you are welcome – Alex Jun 10 '14 at 13:32
  • but is any function, which is $o(\frac1n)$, of the form $\frac{1}{n^{1+\epsilon}}$ ? – OBDA Jun 10 '14 at 13:47
  • take the limit: $ \frac{\frac{1}{n^{1 + \epsilon}}}{\frac{1}{n}}$. What do you get? – Alex Jun 10 '14 at 13:48
  • here of course $\epsilon>0$ is arbitrarily small. Think of the smallest real number possible and take $\epsilon$ less than it. – Alex Jun 10 '14 at 13:50
  • You showed the converse – OBDA Jun 10 '14 at 14:12
  • What do you mean? The limit is $\frac{1}{n^ \epsilon} \to 0$, hence $\frac{1}{n^{1 + \epsilon}} = o(\frac{1}{n})$ – Alex Jun 10 '14 at 14:17
  • I asked: If you have a function $f$ such that, $f=o(\frac1n)$ is $f$ necessarily of the form $\frac{1}{n^{1+\epsilon}}$ for some $\epsilon$, or can $f$ also be for example $\frac{1}{n\log(n)}$ (there might be better examples). How do you justify your answer then ? – OBDA Jun 10 '14 at 14:27
  • plese see the edit – Alex Jun 10 '14 at 15:07